Solved Examples Question 1 Successive discounts of 10% and 20% are equivalent to a single discount of (1) 30% (2) 15% (3) 72% (4) 28% Solution: (4) Let P be the original price. Then a discount of 10% gives a new price P - .1 P = .9 P. Following this by a discount of 20%, we have .9P - .2 (.9P) = .72 P Thus, the net discount is P – .72 P = .28P or 28% Question 2 After selling a watch, Shyam found that he had made a loss of 10%. He also found that had he sold it for Rs 27 more, he would have made a profit of 5%. The initial loss was … (1) Rs 2.70 (2) Rs 16.65 (3) Rs 18.00 (4) Data insufficient Solution: (3) 15% of the cost price of the watch = Rs 27 ∴10% of the cost price of the watch

18Rs15

1027Rs =

×=

Question 3 The value of a share of stock P and the value of a share of stock Q each increased by 16%. If the value of a share of stock P increased by 16% cents and the value of a share of stock Q increased by $1.68, what is the difference between the value of stock Q and the value of stock P before the increases? (1) $8.00 (2) $9.50 (3) $10.00 (4) $10.50 Solution: (2) Note: Only one basic insight is needed to handle the problem. Knowing the amount and percent of an increase is sufficient to allow you to calculate the starting and ending amounts. Here we need the starting amounts. For stock P we know that an increase of 16 cents is equal to 16% of the original value of the stock: $0.16 = 16% of Original Value $0.16 = 0.16 × OV

$0.16OV

0.16=

Original Value = $1.00 For stock Q we know that an increase of $1.68 is equal to 16% of the original value of the stock: $1.68 = 16% of Original Value $1.68 = 0.16 × OV $1.68

OV0.16

=

Original Value = $10.50 Now we find the difference between the original values: $10.50 - $1.0 = $9.50 Directions for Questions. 4 to 6: Refer to the data below and answer the questions that follow. `A, B and C form a partnership investing in the ratio 1 : 2: 3. A agrees to work as the Director for a payment of Rs 500 along with 10% of the total profit as incentive. The partnership is for a period of 3 years. Half the profits after payments to A are out in a corpus fund for investment, e.g., at the end of 1 year, a sum is invested. Similarly at the end of the second year an amount is invested. This fund earns 10% simple interest per annum. The total profit in each year is 50% greater than that of the previous year. At the end of 3 years, the interest accumulates to Rs 240. The total profit less wages and incentive to A is net profit. While half the net profit is saved every year, the other half is shared as per partnership investment. The invested amount is not shared nor the interest shared. Question 4 What is the profit made in the third year? (1) 4000 (2) 4500 (3) 5000 (4) 6500 Solution: (2) Now we get profit in one third year = 2.25x = 4500

Question 5 Find A’s earnings in the 3 year both as Director as well as partner? (1) 340 (2) 2400 (3) 1100 (4) 2600 Solution: (1) A’s total earning

( )6

3550220013001475.01500

++++= x

3405.30375.5879501500 ==++= Question 6 What is the amount earned by C over the 3 years? (1) 6600 (2) 3300 (3) 2200 (4) 1770 Solution: (4) Amount earned by C

( ) 17705.176270504

1

6

3355022001300

2

1≈=×=×++=

DIRECTIONS for questions 7 and 8: Refer to the data below and answer the

questions that follow.

Per capita rice consumption of a country is 40 kg. (per capita = Total consumed/Total population.) Question 7 If only 60% of the population eat rice, find the consumption per rice-eater. (1) 200/3 kg (2) 400/3 kg (3) 60 kg (4) population statistic needed to solve the question

Solution: (1) Let population = N. Total consumption = 40N kg. Number of rice eaters = 0.6N. ∴Consumption per rice eater = 40N/(0.6)N = 400/6 = 200/3 kg.

Question 8 If the number of rice-eaters increases by 20% and total population increases by 15%. while the consumption per rice-eater remains unchanged, find the new per capita rice consumption. (1) 41.7 (2) 48 (3) 54.2 (4) Insufficient data Solution: (1) New number of rice eaters = (0.6)N(1.2) (20% increase) Total population = N1 = (1.15)N New per capita rice consumption (0.6)(1.2N × 400/6 / (1.15)N = 41.7 kg. Hence, (1). Question 9 The income of a person becomes 3 times itself. What is the percentage change in income? (1) 200 % (2) 100 % (3) 50 % (4) 50% Solution: (1) Let the income be Rs. 100. If it becomes 3 times (i.e. Rs. 300), then the percentage change of the income = ((300 - 100)/100)100 = 200 % increase. Question 10 The price of an item increases first by 25 percent and then decreases by x %. If the final price is the same as the initial price, then the value of x in % is (1) 40 (2) 25 (3) 20 % (4) Insufficient data Solution: (3) If the old price = Rs. 100, then the price after the first increase = 125. Then the decrease % required so that it becomes equal to the original = 20 %.

Solved Examples

Question 1

Two high school classes took the same test. One class of 20 students made an

average grade of 80%; the other class of 30 students made an average grade of

70%. The average grade for all students in both classes is:

(1) 75% (2) 74% (3) 72% (4) none of these

Solution: (2)

Suppose the maximum marks for the test were 100. So the class of 20 students

scores a total of 20 80 = 1600 marks, while the class of 30 students scores a

total of 30 70 = 2100 marks. Taken together, the fifty students score a total of

1600 + 2100 = 3700 marks. So, on an average, each of these 50 students scores

3700

50 = 74 marks or 74 % marks (since we have assumed the maximum marks

of the test to be 100).

Question 2

A salesman averaged $23 for each day he was on the road but for two of these

days he made only $27 altogether. Apart form these two bad days, he had

averaged $24 for the other days he worked. How much did the salesman earn

that month?

1) $496 2) $483 3) $474 4) $501

Solution: (2)

Let the number of days in the month be �n�. Total earning in the month = 23n

Total earning for (n � 2) days (n � 2) 24

(n � 2) 24 + 27 = 23n. 24n - 48 + 27 = 23n n = 21

Total earning = 21 23 = $483 Hence, (2).

Question 3

In a class of 17 students, average marks in English were 15 and average marks

in Maths were 24. Average total marks per student in English, Maths and Hindi

are 59. Find the average marks of the class in Hindi.

(1) 18.3 (2) 20 (3) 23 (4) Cannot be

determined

Solution: (2)

Total in English = 17 15; Total in Maths = 17 24

Overall total = 17 59

Total makes in Hindi = 17 59-(17 15 + 17 = 24)

= 17 (59 � 39) = 17 20

Average in Hindi = 20

Directions for Q. 4 and Q. 5: Refer to the data below and answer the questions

that follow.

Three distinct one digit numbers are taken and all the numbers that can be

formed by different combinations of these primes taken all at a time are listed.

The difference between the largest and smallest numbers formed is 495. The

sum of digit is more than 13.

Question 4.

What is the average of the various numbers that can be found?

(1) 518 (2) 777 (3) 463 (4) 1357

Solutions: (1)

6 numbers can be formed.

Average

6

1725725572572275257

752275210275210026

1

5183

1554141401400

3

1

Question 5.

The ages of 13 persons from an A.P in ascending order of magnitude. The

youngest two and the eldest two left the group. What was the change in the

change in the average age?

(1) The average increases

(2) The average decreases

(3) The two averages are equal

(4) cannot be determined

Solution: (3)

If A.P is a � 5d, � a � d, a, a + d, � a + 5d, a + 6d, the old average as well

as the new average is a.

Question 6

A group of students decided to buy a tape - recorder in the range of 170 to 195

rupees. But, at the last moment 2 students backed out of the decision so that the

remaining students had to pay Re. 1 each more than they had planned. What

was the price of the tape recorder if the students paid equal shares?

1) Rs.175 2) Rs.180 3) Rs.185 4) Rs.190

Solution: (2)

Price = Rs. x Number of students = n

Then, the contribution made by each student in the two cases are Rs.x/n and Rs.

x/(n - 2) respectively.

Given, x/(n � 2) � x/n = 1 => x = ((n2 � 2n)/2)

Now, 170 < x < 195 => 170 < ((n2 � 2n)/2) < 195

=> n2 - 2n � 340 > 0 => n > 19.5 and n2 - 2n � 390 < 0 => n < 20.7

=> 19.4 < x < 20.7

But �ii� is a natural number, we have n = 20.

:. Cost of the tape recorder = ((n2 � 2n)/2) = (400 - 40)/2 = Rs.180. Hence, (2).

Question 7.

A milkman buys milk that is 80% pure. He then dilutes it further by adding water.

If he added a litre of water to 4 litres of milk bought at the rate of Rs. 12 per litre,

then what is his profit, assuming that he sells all that he so obtains at the rate of

Rs. 12 per litre?

1) 15% 2) 25% 3) 20% 4) 162/3%

Solution: (2)

Bought at 4 12 = Rs.48, sold at 5 12 = Rs.60 :. Profit = Rs.12

:. Percentage of profit = 12/48 100 =25%. Hence, (2).

Question 8

4 litres of a certain mixture of alcohol and water is at 50 % strength. It is added

with one litre of water. The alcohol strength of the new mixture is:

1) 40% 2) 25% 3) 20% 4) 12.5%

Solution: (1)

4 litres (50%) = 2 litres alcohol. 1 litre, of water is added.

Now 2 litres alcohol in a total of 5 litres;. Alcohol strength = 2/5 100 = 40%.

Question 9

A cup of milk contains 3 parts pure milk and 1 part water. How much of the

mixture must be withdrawn and substituted with water in order that the resulting

mixture may be half milk and half water?

1) ½ 2) 1/3 3) ¼ 4) 1/5

Solution: (2)

Let there be 3x parts pure milk and x parts water originally.

Let the part of the mixture that is withdrawn be y.

Now, the milk left = 3x � y ¾( :. ¾ of y is milk)

Water left = x � y ¼

Now, 3x - y ¾ = x - y ¼ + y

:. 12x -3y = 4x - y + 4y

8x = 6y and y = 8x/6 = 4/3 x

:. y/4x -1/3; y is 1/3rd of the mixture. Hence, (2).

Question 10

Average weight of a student in a class is 43 kg. Four new students are admitted

to the class whose weights are 42 kg, 36.5 kg, 39 kg. and 42.5 kg. respectively.

Now their average weight is 42.5 kg. Find the number of students in class in the

beginning.

1) 10 2) 15 3) 20 4) 25

Solution: (3)

Let the number of students originally be �n�. If the average weight is 43, total

weight of all students is 43n.

Now (43n + 42 + 36.5.+ 39 + 42.5)/ n + 4 = 42.5

43n + 160 = 42.511 + 170; :. 0.5n = 10; :. n = 20.

Solved Examples

Question 1

Rs. 1080 was divided among Arun, Bhaskar and Cheenu in a certain ratio. Had

each of them received Rs. 6 less than their actual share, then for every part Arun

got, Cheenu would have got 2 parts and for every 2 parts Arun got, Bhaskar

would have got 3 parts. How much was Bhaskar�s actual share?

(1) Rs. 354 (2) Rs. 472 (3) Rs. 236 (4) none of these

Solution: (4)

Total = Rs. 1080 .

When each share is decreased by Rs. 6, total reduction = 6 × 3 = Rs. 18

new total to which the ratios corresponds = 1080 � 18 = 1062.

Given that, A : C = 1 : 2 and A : B = 2 : 3 A : B : C = 2 : 3 : 4

B = (3/9) × 1062 = 354 B�s total share = 354 + 6 = Rs. 360.

Question 2

The ratio of income of Anuj and Bharat and also those of Bharat and Champak

are in the ratio of 2 : 3. A third of Champak�s income exceeds half of Anuj�s

income by Rs. 80. If each of them spend the same amount of money, then their

savings are in the ratio of 1 : 9 : 21. What is their combined expenditure?

(1) Rs. 300 (2) Rs. 280 (3) Rs. 21 (4) none of these

Solution: (4)

A : B = 2 : 3 and B : C = 2 : 3 A : B : C = 4 : 6 : 9

Let incomes = 4k, 6k, 9k. given that 1/3(9k) � ½(4k) = 80

k = 80

incomes are A = 4k = 320, B = 6k = 480, C = 9k = 720

Let p be the expenditure.

Now, 320 � p; 480 � p; 720 � p be in the ratio of 1 : 9 : 21

9

1

p480

p320 2880 � 9p = 4800 � p

2400 = 8p

p = 300

total spending = 300 × 3 = 900.

Question 3

In a class, there are some boys and girls. The ratios of girls to boys, when 35

boys are taken out and when 5 boys are added, the reciprocals of one another, if

the number of girls doubles, then the ratio of girls to boys becomes 3 : 4. What is

the strength of the class?

(1) 40 (2) 55 (3) 45 (4) 33

Solution: (2)

G/(B � 55) = x/y (assume); G/(B + 5) = y/x G/(B � 35) = (B + 5) /G

G2 = (B + 5) (B � 35) G2 = B2 � 30B � 175 ���(1)

Given that 2G/B = ¾ G/B = 3/B ���.(2)

Put (2) in (1)

0175B3064

B55175B30B

64

B92

22

.

By solving we get B = 40 and G = 3B/8= 15 strength = 55.

Question 4

In a farm which has only cocks and bulls, total count of legs was 12 less than 4

times the total count of heads. How many legs are counted in total?

(1) 6 (2) 12 (3) 3 (d) cannot be determined

Solution: (4)

Let the number of cocks = C which has 1 head, 2 legs ; Bull has B = 1 head, 4

legs

(2C + 4B) + 12 = 4(C + B) ; 2C + 4B + 12 = 4C + 4B

2C = 12 C = 6. legs = 2C + 4B. Since, we cannot know the number of

bulls, we cannot find the value.

Question 5

If p potters working p hours per day for p days produce p pots then how many

pots can be made by q potters working q hours a day for q days?

(1) p3/q3 (2) q3/p2 (3) p3/q2 (4) p2/q3

Solution: (5)

2

222

1

111

W

HDM

W

HDM

2

3

22

2 p

qW

ppp

qqqpW

W

qqq

p

ppp

Question 6

Two numbers are such that their differences, the sum and their produce are in

the ratio of 1:5:12. Find the difference of squares of the numbers.

(1) 5 (2) 15 (3) 10 (4) 20

Solution: (4)

Let the numbers be x, y

(x � y) : (x + y) : xy = 1 : 5 : 12

1

5

yx

yx

x/y = 6/4 = 3/2 ��..(1)

12yx

xy ���(2)

xy = 12(x � y)

(3y/2) y = 12(3y /2 � y) 3y2/2 = 6y y = 4

x = 3y/2 = 6 x2 - y2 = 36 � 16 = 20.

Question 7

If Rs. 1980 is to be distributed to three men Amy, Bunty and Castor in such a

way when Rs. 4, Rs. 5 and Rs. 11 are added to their respective shares, the ratio

becomes 4 : 5 : 11. What is Bunty�s share?

(1) Rs. 500 (2) Rs. 495 (3) Rs. 505 (4) Rs. 490

Solution: (2)

The total amount = 1980 + 4 + 5 + 11 = 200

Share of Bunty in Rs. 2000 = (5/20) × 2000 = Rs. 500

Bunty�s share before adding Rs. 5 = 500 � 5 = Rs. 495.

Question 8

In two alloys, copper and tin are mixed in the ratio 5 : 1 and 1 : 3. If 24 kg of the

first alloy, 32 kg of the second alloy and some pure copper are melted together,

then a new alloy is formed in which the ratio of copper to tin is 2 : 1. Find the

weight of the new alloy.

(1) 76kg (2) 84kg (3) 88kg (4) 104 kg

Solution: (2)

24kg of first alloy has (5/6) (24) = 20kg of copper and (1/6) 24 = 4kg of tin. 32 kg

of second alloy has (1/4) (32) = 8 kg of copper and (3/4) (32) = 24 kg of tin.

If the weight of pure copper melted along with these two alloys is x, then

(20 + 8 + x) : (4 + 24) = 2 : 1 x = 28. Weight of new alloy = 24 + 28 + 28 =

84kg.

Question 9

In Delhi, an Ambassador car with only the driver inside moves at a speed of

30km/hr. Its speed gets reduced by a quantity which is directly proportional to the

number of passengers seated inside. The speed of the car reduces by 10km/hr, if

there are 5 passengers. A minimum of how many passengers should be seated

such that the car does not move at all?

(1) 8 (2) 15 (3) 10 (4) none of these

Solution: (2)

If R is the reduction in speed and N is the number of passengers, then R N

R = NK, where K is the constant of proportionality. Given data, R = 10, when N =

5, hence K = 2 and R =2N. The equation for speed is S = 30 � R S = 30 � 2N.

Car should not move S = 0 30 = 2N N = 15.

Question 10

What number should be subtracted from each of 42, 61, 18, 25 such that the

remainder are proportional?

(1) 3 (2) 2 (3) 1 (4) none of these

Solution: (4)

Say x is the number, x25

x18

x61

x42

(42 � x) (25 � x) = (18 � x) (61 � x). Trying out options we find none of them

satisfy but x = 4 satisfies.

Solved Examples Question 1 If the sum of the cube of a number x and the cube of its reciprocal is 3 times the sum of the number and its reciprocal, find x + 1/x.

(1) 5/2 (2) 6 (3) 83/2 (4) 2 3 Solution: (2) x3 + 1/x3 = 3(x + 1/x) (x +1/x)3 = x3 + 1/x3 + 3(x + 1/x) (x + 1/x)3 = 6(x + 1/x)

(x + 1/x)2 = 6 => x + 1/x = 6 Hence, (2). Question 2 When asked to write (a2 + b2) (x2 + y2) as a sum of two perfect squares, a student gave the following answers: (1) (ax + by)2 + (bx + ay)2 (2) (ax � by)2 + (bx � ay)2 (3) (ax + by)2 + (bx � ay)2 (4) (ax � by)2 + (bx + ay)2 Solution: (4) (a2 + b2) (x2 + y2) = (a2x2 + b2y2) + (b2x2 + a2y2)

= (a2x2 + b2y2 + 2abxy) + (b2x2 + a2y2 � 2abxy) = (ax + by)2 + (bx � ay)2

Also, (a2 + b2) (x2 + y2) = (a2x2 + b2y2 � 2abxy) + (b2x2 + a2y2 + 2abxy)

= (ax � by)2 + (bx + ay)2 Thus (a2 + b2) (x2 + y2) = (ax + by)2 + (bx � ay)2 or (ax � by)2 + (bx + ay)2

Question 3

If 6a b c

a b b c c a, find the value of

b c a

a b b c c a

(1) 0 (2) 1 (3) �1 (4) 3 Solution: (4)

1 1 1a b c

a b b c c a3

3b c a

a b b c c a

Question 4

The sum of two positive numbers is 1

21 , and the sum of their reciprocals is 3. Find

the numbers.

(1) 1 and 1

2 (2)

1

41 and

1

4 (3)

7 and

8

5

8 (4)

3

4 and

3

4

Solution: (3) Let the numbers be x and y

x + y = 1

21

1 1

3x y

1 1

3x y

x y xy

3

32xy

1

12xy

1

2xy

(x + y)2 � 4xy = (x � y)2

292 ( )

4x y

21

4x y

1

2x y

x + y = 1

21 x + y =

1

21

1

2

2 2

x y

x

1

2

2 1

x y

x

x = 1 1

2x

and y = 1

2 y = 1

The numbers are 1 and 1

2. (Both give positive answers).

Question 5 Find the value of K if (x + 1) is a factor of x8 + Kx3 � 2x + 1. (1) 4 (2) 1 (3) 2 (4) 3 Solution: (1) p(x) = x8 + K(x)3 - 2x + 1, p(-l) = (-1)8 + K(-1)3 - 2(-1) + 1 = 0 1 - K + 2 + 1 = 0. K = 4 Hence, [1]. Question 6 If a + 2b =10 and ab = 15, find the value of a3 + 8b3. (1) 10 (2) 50 (3) 100 (4) 150 Solution: (3) We will use the identity

In this case, x = a, y = 2b

xyxyyxyx 3333

babababa 22382 333

100101561026283333 baabbaba

Question 7 Reduce

22

to its simplest form, given that , and are all equal. (1) 0 (2) 1 (3) 4 (4) 5 Solution: (2)

22

2

2 2

1122

2

22

Question 8

kxxgkxxf findf,oggofIf.23,1

(1) 1 (2) 3 (3) 2 (4) Indeterminate Solution: (3)

2and42

233123iii

ii2311

)i(12323

kk

kxkkx

kxkxgxfgfgo

kxxfxgfgof

Question 9 If the value of a4 + 1/a4 = 119. the value of a3 - 1/a3 is: (1) 27 (2) 36 (3) 45 (4) 54

Solution: (2)

(a4 + 1/a4) = 119

(a4 + 2 + 1/a4) = 121 :. (a2 + 1/a2)2 = 112 :. a2 + 1/a2 = 11

:. (a2 � 2 + 1/a2) = 11-2 :.(a � 1/a)2 = 9 :. a-1/a = 3 Now, (a � 1/a)2 = (3)3 = a3 � 1/a3 � 3a2. 1/a + 3ax1/a2

:. 27 = (a3 � 1/a3) � 3(a � 1/a) :. 27 = (a3 �1/a3) � 3(3) :. a3 � 1/a3 = 36. Hence, (2). Question 10 If f(x) = 1/x + 1, then f{ f( f( f(x)))} = ? (1) 2x + 3/3x + 5 (2)3x + 2/5x + 3 (3) 2x - 3/5x + 3 (4) None of these Solution: (1) f(x) = 1/x +1 . f(1/x + 1) = 1 / 1/(x + 1) +1 =x +1/ x + 2 f(x + 1 / x +2) = 1 / (x + 1/x +2) + 1 = x +2/ 2x + 3 f(x + 2/2x +3) = 1 / (x + 2/2x + 3) + 1 = 2x +3/ 3x + 5 Hence, (1)

Solved Examples

Question 1

What is the numerical difference between the perimeter and the area of a

rectangle whose sides are the roots of the equation 01x22x32 ?

(1) 2 (2) 3 (3) 5 (4) 7

Solution: (3)

Roots of the equation 12232

xx

Squaring, 3x2 � 2 = 4x2 + 1 � 4x

x2 � 4x +3 = 0 x = 1, x = 3

area = 3 × 1 = 3; perimeter = 2(3 + 1) = 8

difference = 8 � 3 = 5.

Question 2

and are distinct roots of a quadratic equation where 2 + 1 = 3 and 2 = 3

� 1. Find the equation whose roots are [(1/ ) + 2] and [(1/ ) + 2].

(1) x2 � 7x � 11 = 0 (2) x2 + 7x + 11 = 0

(3) x2 � 7x + 11 = 0 (4) x2 + 7x � 11 = 0

Solution: (3)

It is given that 2 - 3 + 1 and 2 - 3 + 1 = 0

both and satisfy the equation x2 � 3x + 1 = 0

+ = 3 and = 1

the new roots are (1/ ) + 2 and (1/ ) + 2

sum = 74411

product of the roots = 114221

the required equation is x2 � 7x + 11 = 0.

Question 3

A standard quadratic equation has a = 5, b = -4 and c = 6. Find the difference

between the product of the roots and the sum of the roots.

(1) 0.4 (2) 0.8 (3) 0.6 (4) 1

Solution: (1)

The product of the roots = c/a = 6/5 = 1.2

The sum of the roots = - b/a = (4/5) = 0.8

Thus the difference = 0.4

Question 4

The roots of the quadratic equation ax2 + bx + c = 0 are integers and differ by 2.

The result obtained by multiplying the sum of the roots and the product of the

roots is always divisible by

(1) 9 (2) 12 (3) 18 (4) 24

Solution: (2)

let the roots be and + 2. Sum of roots = 2 + 2 = 2( + 1) ��.(1)

Product of roots = ( + 2) ���..(2)

(1) × (2)

= 2 ( + 1) ( + 2) and this is divisible by 12

since, , + 1 and + 2 are 3 consecutive numbers and their product is

divisible by 6.

Question 5

2x2 � 3x + k = 0 and x2 � 5x � 3k = 0 have a common root for some value of k.

Find the common root.

(1) - ½ (2) 2 (3) 1 (4) 3/2

Solution: (2)

Let be the common root. shall satisfy each of the equations

2 2 - 3 + k = 0 ���(1) and 2 - 5 - 3k = 0 ���(2)

(1) � 2 × (2) gives the equation 7 + 7k = 0

= - k.

substituting in (1)

2 2 - 3 + (- ) = 0; 2 2 - 4 = 0 ; 2 ( - 2) = 0

= 0 or 2

Question 6

Find the ratio of the LCM and the HCF of the roots of the equation 8x2 � 22x + 15

= 0.

(1) 30 : 1 (2) 15 : 2 (3) 15 : 4 (4) 60 : 1

Solution: (1)

Equation is 8x2 � 22x + 15 = 0

Roots = 16

20,

16

24

16

222

16

48048422

Roots are 3/2, 5/4

LCM(3/2, 5/4) = 2

15

4,2HCF

5,3LCM

HCF (3/2, 5/4) = 4

1

4,2LCM

5,3HCF30

4/1

2/15

HCF

LCM

Question 7

The roots of the equation x2- 2 3x + 3 = 0 are:

1) rational and unequal 2) irrational and unequal

3) rational and equal 4) real and equal

Solution: (1)

b2 - 4ac = (-2 3)2-- 4(1)(3) = 12 - 4(3) = 0. Since = b2 - 4ac = 0, the roots are real

and equal. Hence, (4).

Question 8

The equation 2

1

1

x bx m

ax c m has roots which are equal in magnitude but opposite

in sign. Therefore m = ?

(1) 1 (2) a b

a b (3)

a b

a b (4) -1

Solution: 3

(x2 - bx) (m + 1) - (ax � c) (m - 1) = 0

x2(m + 1) - bx m - ax(m - 1) - bx + c(m - 1) = 0;

x2(m + 1) - x(bm + am - a + b) + c(m - 1) = 0

If roots are equal and opposite than, (bm + am - a + b) = 0

bm + am = a - b; m = a b

a b Hence, [3].

Question 9

Solve : ((x + 3) / (x + 4)) � (x � 1) / (2x � 1)) = ½

(1) 2.5 (2) 3 (3) 1.5 (4) 2

Solution: (4)

((x + 3)/(x + 4)) � ((x � 1) / (2x � 1)) = ½ :. (x + 3)/(x + 4) = ½ + (x � 1)/(2x-1);

:. (x + 3)/ (x + 4) = (2x - 1 + 2x � 2)/(4x � 2)

:. (x + 3)(4x - 2) = (4x - 3)(x + 4)

:. 4x2 + 12X � 2X � 6 = 4x2 + 16x - 3x - l2 . .

:. 10x � 6 = 13X -12 :. X = 2.Hence,(4).

Alternatively,

The values could also be checked by substituting the values from the choices.

Question 10

The roots of the equation 12x2 + mx + 5 = 0 will be in the ratio 3 : 2 if m equals:

1) 1/12 2) 5/12 3) 5 10 4) 5/12 10

Solution: (3)

Sum of the roots = + = -m/12

Product of the roots = = 5/12

/ = 3/2 :. 2 = 3 or = 3/2

Now, 3/2 2 = 5/12 :. 2 = 5/18 ��(i)

Also 3/2 + = -m/12 :. = -m/30 ...(ii)

m2/ 9000 = 5/18 :.m2 = 5x900/18 =250 :. m = 5 10. Hence, (3).

Solved examples

Question 1

A cake weighing 10g is cut into three unequal pieces. The largest piece was twice the

weight of the middle sized piece and the smallest piece is 50gms lighter than the largest

piece. Find the weight of the smallest piece.

(1) 10gms (2) 15gms (3) 20gms (4) 25gms

Solution: (1)

If the weight of the smallest piece is x gm, then the weight of the largest piece is (x + 50)

gm and the middle sized piece has a weight of 2

50xgm. Then,

2x + 50 + 2

50x= 100 x = 10gm.

Question 2

Let R = gS � 4. When S = 8, R = 16. When S = 10, R is equal to

(1) 11 (2) 14 (3) 20 (4) 21

Solution: (4)

From the given conditions,

16 = 8g � 4

g = 2 ½

R = (2½) 10 � 4 = 21

Question 3

115. For the simultaneous equation 2x � 3y = 8, 6y � 4x = 9

(1) x = 4, y = 0 (2) x = 0, y = 3/2

(3) x = 0, y = 0 (4) there is no solution

Solution: (4)

When the two given equations are plotted, they represent two parallel lines. Hence they

never meet. This in turn means that there is a infinite of set of answers which would satisfy

the equations.

Algebraically, if a1x + b1y = c1 and a2x + b2y = c2 then, multiply the first equation

by b2, the second by b1 and subtracting we have

(a1b2 � a2b1) x = c1b1 � c2 b1 1221

1221

baba

bcbcx

thus, if a1b2 � a2b1 = 0 and c1 0, c2 0, x is undefined and no solution exists.

Question 4

From a group of boys and girls, 15 girls leave. There are then left two boys for each girl.

After this 45 boys leave. There are then 5 girls for each boy. The number of girls in the

beginning was

(1) 40 (2) 43 (3) 29 (4) 50

Solution: (1)

If B is the original number of boys and G the original number of girls, then

5

1

15G

45B,2

15G

B G = 40

Question 5

If 101x + 202 y = 1313; 202 x + 101 y = 1111, then value of x and y is

(1) x = 3, y = 6 (2) x = 6, y = 3

(3) x = 5, y = 3 (4) x = 3, y = 5

Solution: (4)

101x + 202 y = 1313 ����.(1)

202x + 101y = 1111 ����.(2)

adding (1) and (2),

303 (x + y) = 2424 x + y = 8 ����.(3)

subtracting (2) from (1), 101 (y � x) = 202

y � x = 2 ����(4)

adding (3) and (4), we get 2y = 10 y = 5 ���..(5)

subtracting y = 5 in Eq. (3) we get x = 8 � 5 = 3

Alternatively, one can put the values in the parent equations and check which of the pair of

values satisfy the equations

Question 6

Five years ago, the age of a mother was four times the age of her daughter. Fifteen years

hence, the mother�s age will be twice the age of the daughter. Find the present age of the

daughter.

(1) 24 years (2) 20 years (3) 18 years (4) 15 years

Solution: (4)

Let the present age of the mother and the daughter be M and D respectively ten

M � 5 = 4 (D � 5) ���..(1)

M + 15 = 2 (D + 15) ���...(2)

Eq. (2) - Eq. (1) we get

�20 = 2D � 50 D = 15 years .

Question 7

The excess of a two � digit number over the number formed by reversing its digits is 18. If

the sum of its digits is 4 times the difference of the digits, find the number.

(1) 86 (2) 75 (3) 64 (4) 53

Solution: (4)

Let the two digit number be 10a + b. From the condition of the problem, we get

a + b = 4(a � b) 4 (a � b) or 4(b � a). As a > b, a + b = 4(a � b) a/b = 5/3. Hence,

a must be a multiple of 5. As a is a single digit integer, a = 5. Hence, b = 3.

Question 8

The total age of Amar and Bhawan 10 years ago, and the total of their present ages are in

the ratio of 2 : 3. If the ratio of their present ages is 3 : 1, find the present age (in years) of

Amar.

(1) 27 (2) 24 (3) 21 (4) 45

Solution: (4)

Let the present age of Amar and Bhawan be A and B then,

A � 10 + B � 10 = 2/3 (A + B) A + B = 60

Hence, A = ¾ (60) = 45.

Question 9

Instead of multiplying a number with 5/6, Ajay multiplied it with its reciprocal. As a result, he

got an answer which is 242 more than the expected result. Find the number.

(1) 660 (2) 315 (3) 420 (4) 840

Solution: (1)

Let the number be x. Ajay was expected to find 5x/6. He found 6x/5. Their difference is

given by

242x30

11242x

6

5x5

6 x = 660.

Question 10

The amount of money with Ramesh and Raj have, is in the ratio of 7 : 4. If Ramesh has Rs.

12 more than Raj, find the total amount with both.

(1) Rs. 44 (2) Rs. 55 (3) Rs. 66 (4) Rs. 77

Solution: (1)

Let the amount of money with Ramesh and Raj be 7x and 4x respectively. Then

7x � 4x = 3x = 12 x = 4

total amount with both = 11x = Rs. 44.

Solved examples

Question 1

A cake weighing 10g is cut into three unequal pieces. The largest piece was twice the

weight of the middle sized piece and the smallest piece is 50gms lighter than the largest

piece. Find the weight of the smallest piece.

(1) 10gms (2) 15gms (3) 20gms (4) 25gms

Solution: (1)

If the weight of the smallest piece is x gm, then the weight of the largest piece is (x + 50)

gm and the middle sized piece has a weight of 2

50xgm. Then,

2x + 50 + 2

50x= 100 x = 10gm.

Question 2

Let R = gS � 4. When S = 8, R = 16. When S = 10, R is equal to

(1) 11 (2) 14 (3) 20 (4) 21

Solution: (4)

From the given conditions,

16 = 8g � 4

g = 2 ½

R = (2½) 10 � 4 = 21

Question 3

115. For the simultaneous equation 2x � 3y = 8, 6y � 4x = 9

(1) x = 4, y = 0 (2) x = 0, y = 3/2

(3) x = 0, y = 0 (4) there is no solution

Solution: (4)

When the two given equations are plotted, they represent two parallel lines. Hence they

never meet. This in turn means that there is a infinite of set of answers which would satisfy

the equations.

Algebraically, if a1x + b1y = c1 and a2x + b2y = c2 then, multiply the first equation

by b2, the second by b1 and subtracting we have

(a1b2 � a2b1) x = c1b1 � c2 b1 1221

1221

baba

bcbcx

thus, if a1b2 � a2b1 = 0 and c1 0, c2 0, x is undefined and no solution exists.

Question 4

From a group of boys and girls, 15 girls leave. There are then left two boys for each girl.

After this 45 boys leave. There are then 5 girls for each boy. The number of girls in the

beginning was

(1) 40 (2) 43 (3) 29 (4) 50

Solution: (1)

If B is the original number of boys and G the original number of girls, then

5

1

15G

45B,2

15G

B G = 40

Question 5

If 101x + 202 y = 1313; 202 x + 101 y = 1111, then value of x and y is

(1) x = 3, y = 6 (2) x = 6, y = 3

(3) x = 5, y = 3 (4) x = 3, y = 5

Solution: (4)

101x + 202 y = 1313 ����.(1)

202x + 101y = 1111 ����.(2)

adding (1) and (2),

303 (x + y) = 2424 x + y = 8 ����.(3)

subtracting (2) from (1), 101 (y � x) = 202

y � x = 2 ����(4)

adding (3) and (4), we get 2y = 10 y = 5 ���..(5)

subtracting y = 5 in Eq. (3) we get x = 8 � 5 = 3

Alternatively, one can put the values in the parent equations and check which of the pair of

values satisfy the equations

Question 6

Five years ago, the age of a mother was four times the age of her daughter. Fifteen years

hence, the mother�s age will be twice the age of the daughter. Find the present age of the

daughter.

(1) 24 years (2) 20 years (3) 18 years (4) 15 years

Solution: (4)

Let the present age of the mother and the daughter be M and D respectively ten

M � 5 = 4 (D � 5) ���..(1)

M + 15 = 2 (D + 15) ���...(2)

Eq. (2) - Eq. (1) we get

�20 = 2D � 50 D = 15 years .

Question 7

The excess of a two � digit number over the number formed by reversing its digits is 18. If

the sum of its digits is 4 times the difference of the digits, find the number.

(1) 86 (2) 75 (3) 64 (4) 53

Solution: (4)

Let the two digit number be 10a + b. From the condition of the problem, we get

a + b = 4(a � b) 4 (a � b) or 4(b � a). As a > b, a + b = 4(a � b) a/b = 5/3. Hence,

a must be a multiple of 5. As a is a single digit integer, a = 5. Hence, b = 3.

Question 8

The total age of Amar and Bhawan 10 years ago, and the total of their present ages are in

the ratio of 2 : 3. If the ratio of their present ages is 3 : 1, find the present age (in years) of

Amar.

(1) 27 (2) 24 (3) 21 (4) 45

Solution: (4)

Let the present age of Amar and Bhawan be A and B then,

A � 10 + B � 10 = 2/3 (A + B) A + B = 60

Hence, A = ¾ (60) = 45.

Question 9

Instead of multiplying a number with 5/6, Ajay multiplied it with its reciprocal. As a result, he

got an answer which is 242 more than the expected result. Find the number.

(1) 660 (2) 315 (3) 420 (4) 840

Solution: (1)

Let the number be x. Ajay was expected to find 5x/6. He found 6x/5. Their difference is

given by

242x30

11242x

6

5x5

6 x = 660.

Question 10

The amount of money with Ramesh and Raj have, is in the ratio of 7 : 4. If Ramesh has Rs.

12 more than Raj, find the total amount with both.

(1) Rs. 44 (2) Rs. 55 (3) Rs. 66 (4) Rs. 77

Solution: (1)

Let the amount of money with Ramesh and Raj be 7x and 4x respectively. Then

7x � 4x = 3x = 12 x = 4

total amount with both = 11x = Rs. 44.

Solved Examples

Question 1

If ax = cq = b, then cy = az = d, then

(1) xy = qz (2) z

q

y

x (3) x + y = q + z (4) x � y = q � z

Solution: (1)

Since, c2 = ax, c = az/ y

qzxyy

zqxadc zqy/zq

Question 2

Arrange the following numbers in increasing order:

I.

33

4 II.

22

5 III. (0.3)2 IV. (-1.2)2

(1) I, II, III, IV (2) II, III, I, IV (3) I, III, II, IV (4) IV, II, III, I

Solution: (3)

I

33 2

4 6

7

4

II

22 4

5 20.16

5

III (0.3)2 = 0.09

IV (- 1.2)2 = 1.44

Hence I < III < II < IV.

Question 3

If , then find the value of x. 633 5x4x

(1) 2 (2) � 2 (3) � 3 (4) � 4

Solution: (3)

633 5x4x

3x . 34 � 3x . 35 = � 6

3x . 34(1 � 3) = � 6

3x = 3. 3-4

3x = 3-3

x = � 3.

Question 4

If , then find the value of m, given x 1 and x 0. 43m3 xx

(1) 4 (2) 3 (3) 9 (4) 27

Solution: (4)

43m3 xx

3 × m = 3 × 3 × 3 × 3

m = 27.

Question 5

Arrange in ascending order of magnitudes

I. 2 II. 3 III. 3 4 5

(1) III, II, I (2) II, III, I (3) I, III, II (4) I, II, III

Solution: (4)

I = 1

22

II = 1

33

III = 1

45

Raise each to the 12th power, 12 being the L.C.M. of 2, 3, 4

121

622 2 64

121

433 3 81

121

345 5 125

the order is I, II, III

Question 6

The value of

44/3 0.5 0.751/ 2 34 . 64 . 0.25 . 16 is

(1) 8 (2) 4 (3) 16 (4) 1

4

Solution: (3)

Given

4/31/ 2

4 /3 3/ 41/ 2 3 21

4 4 . 44

1/ 2 4/3 3 4/3 4/3 2

1/ 2 4/3

14 4

4

2/3 16/3 2/3 2 14/3 8/34 .4 .4 .4 4 .4

3/ 4 4/3

4

6/34

24 16

Question 7

If ax = by = cz and b = ac, then �y� equals:

(1) 2xz

x z (2)

xz

x z (3)

2

xz

x z (4)

xz

z z x

Solution: (2)

Let ax = by = cz = k, then a = 1

xk, b =

1

yk, c =

1

zk

b = ac

=

1 1 11

y x xzk k k k1

z

1 1 1

y x z

1 x z

y xz

y = xz

x z

Question 8

If 3 1/ 43 5a and 4 5b 3 , then the value of ab is

(1) 3 (2) 3

4 (3) 9 (4)

3

2

Solution: (4)

Given 1 31

4 433 5 or3a a 5

41 11

22 245 3 3 or5 3b b 3

Combining

32 34

23 3 3a b b

Equating the indices, 3 3or

2 2a ab

b.

Question 9

If 31 . 32 . 33 � 3X = 1/(9)-5 the value of x is:

(1) 10 (2) 6 (3) 5 (4) 4

Solution: (4)

31 . 32 . 33 ...3x = 1/9-5

i.e. 3(1+2+3+�.x) = 1/((3)2)-5

3 x(x+1)/2 = 310

x(x+1)/2=10

x2+x-20 = 0

x = 4 or x = - 5

x = 4 .

Question 10

If , find the value of x. 5122x3

(1) 2 (2) 3 (3) 4 (4) 5

Solution: (1)

5122x3

93 22

x

x = 2

Solved Examples

Question 1

If + is represented by -, - by , by , and by +, then the value of the following

expression is �

112823547

(1) 0 (2) 1 (3) 2 (4) 3

Solution: (1)

The expression can be written as {[{(7 + 4) � 5} 3] × 2 + 8} 12 � 1 = 0.

Question 2

If the sum of the first �n� natural numbers is 171, find the sum of the cubes of the

first n natural numbers.

(1) 29, 241 (2) 28, 421 (3) 30, 141 (4) 29, 421

Solution: (1)

The sum of the first n natural numbers is 1

1712

n n

The sum of their cubes is

2

21

171 29,2412

n n

Question 3

Multiply 143325 by a suitable least number to make it a perfect square and find

its square root.

(1) 1365 (2) 1225 (3) 1305 (4) 1215

Solution: (1)

143325 = 52 × 32 × 72 × 13

If this is multiplied by 13, it becomes a perfect square

52 × 32 × 72 × 132

The square root of the above expression is 5 × 3 × 7 × 13 = 1365

Question 4

What must be added to the product (a � 1) (a + 1) (a + 3) (a + 5), to make it a

perfect square?

(1) 9 (2) 16 (3) 25 (4) 10

Solution: (2)

(a � 1) (a + 5) (a + 1) (a + 3) = (a2 + 4a � 5) (a2 + 4a + 3)

= p (p + 8), where, p = (a2 + 4a - 5)

= p2 + 8p

If 16 be added, then p2 + 8p + 16 becomes (p + 4) 2

Question 5

What is the lowest number which when divided separately by 15, 20, 36 and 48,

will in each case leave 9 as the remainder?

(1) 360 (2) 720 (3) 729 (4) 529

Solution: (2)

L.C.M. = 32 × 42 × 5 = 9 × 16 × 5 = 720

Since it leaves 9 as remainder, the required lowest number is 720 + 9 = 729

Question 6

Find the product 50.02 × 19.98.

(1) 2498.996 (2) 2499.9996 (3) 2499.96 (4) 2499.0006

Solution: (2)

50.02 × 49.98 = (50 + 0.02) (50 � 0.02)

= 502 � (0.02)2

= 2500 � 0.0004

= 2499.9996

Question 7

Which is closest to 1?

(1) 2

2 0.02 (2)

2

2

2 0.02 (3)

22 0.2

2 (4)

2

2 0.2

Solution: (2)

(1) 2

0.992.02

nearly

(2) 2

0.992.0004

, because 2.0004 < 2.02

(3) 2

0.54.84

(4) 2

0.992.2

because 2.2 > 2.02

(2) is closest to 1.

Question 8

If a, b, c, d are four consecutive natural numbers, what must be added to the

product abcd to make it a perfect square?

(1) 9 (2) 4 (3) 1 (4) 0

Solution: (3)

Let the numbers be a, a + 1, a + 2 and a + 3

Product = a (a + 1) (a + 2) (a + 3) = a (a + 3) (a + 1) (a + 2)

= (a2 + 3a) (a2 + 3a + 2)

= p (p + 2), putting p = a2 + 3a

= p2 + 2p

To make this a perfect square we must add 1.

Question 9

The product of a two-digit number by the number of the same digits written in the

reverse order is 2430. Find the sum of the squares of the two digits.

(1) 43 (2) 41 (3) 61 (4) 65

Solution: (2)

The two digit number must be either 25 or 45 or 65 or 85; for then only the unit

digit in the product will be zero.

45 will agree as 45 × 54 = 2430

required sum = 42 + 52 = 41

Question 10

A number when divided by a certain divisor leaves a remainder 19; when twice

the same number is divided by the same divisor, the remainder is 13. Then the

divisor is�

(1) 25 (2) 26 (3) 27 (4)28

Solution: (1)

Let the dividend be N and the divisor x. The remainder is 19.

N = qx + 19, (19 < x)

2N = 2qx + 38

When 2qx + 38 is divided by x, we get the quotient 2x + a number which must

contain x once and 13.

x + 13 = 38

x = 25

Solved Examples

Question 1

The number x and y have prime factors as 2 and 3 only and their ratio y

x is 36.

Which of the statements below is true?

(1) The HCF is 36 (2) The LCM is x

(3) The HCF is x (4) The HCF is y

Solution: (3)

Since x is a factor of y, HCF of the two numbers is x and LCM is y.

Question 2

Four bells are heard at intervals of 5, 8, 10, 12 seconds respectively. How many

times can they be heard within 8 minutes of time, excluding the initial one?

(1) 8 (2) 6 (3) 12 (4) 4

Solution: (4)

L.C.M of 5, 8, 10 and 12 is 120.

Therefore, at the end of every 120 seconds the bells will be heard

simultaneously.

8 minutes = 480 seconds, therefore, the bells will be heard simultaneously 4

times (480/ 120 = 4).

Question 3

The LCM of two numbers is 693 and their HCF is 7. If the sum of the two

numbers is 140, the numbers are

(1) 84, 56 (2) 105, 35 (3) 91, 49 (4) 77, 63

Solution: (4)

Let one of the numbers be x. The other number is 140 � x.

Product of 2 numbers = their (L.C.M) × (H.C.F)

i.e. x (140 � x) = 693 × 7

-x2 + 140x = 4851

x2 � 140x + 4851 = 0

(x � 77) (x � 63) = 0 giving x = 77 or 63

Therefore, the numbers are 77 and 63.

Question 4

What is the greatest number of three digits which when divided by 3, 4, 5, 6 and

10 leave the same remainder 2?

(1) 952 (2) 962 (3) 982 (4) 948

Solution: (2)

The L.C.M of 3, 4, 5, 6, 10 is 60. The greatest multiple of 60 and of three digits is

960 i.e. 960 = Multiple of 60

M (3, 4, 5, 6, 10).

The required number is 962.

Question 5

Find the greatest number that will divide 1336, 1080, 1120, so as to leave

remainders 13, 9, 7 respectively.

(1) 17 (2) 21 (3) 23 (4) 37

Solution: (2)

Find the H.C.F of

(1336 � 13); (1080 � 9) and (1120 � 7)

i.e., 1323; 1071 and 1113

The HCF is 21

Question 6

Find the greatest number which will divide 516, 589 and 808 leaving the same

remainder in each case.

(1) 73 (2) 67 (3) 59 (4) 77

Solution: (1)

Since the remainders are the same, the required divisor will exactly divide the

difference of any two of the given three numbers.

The number is the H.C.F of (589 � 516) and (808 � 516), which can be

computed as 73.

Question 7

Write the fraction 301

645in its simplest form.

(1) 7

15 (2)

13

25 (3)

17

35 (4)

19

40

Solution: (1)

Where you cannot give the factor common to the numerator and the

denominator, find their H.C.F.

H.C.F. of 301 and 645 is 43. The fraction is 43 7 7

43 15 15

Question 8

The L.C.M. and H.C.F. of two numbers are respectively 544 and 369. The

difference of the numbers is

(1) 1925 (2) 1980 (3) 2020 (4) 1895

Solution: (2)

5544 = 32 × 23 × 7 × 1 and 396 = 32 × 22 × 11

If Ha and Hb are the numbers with H.C.F. as H = 396, then L.C.M. = Hab = 5544

Hab 5544

H 396, giving ab = 14 or 2 × 7

Taking a as 7 and b as 2, the numbers are 2772 and 792. Their difference is

1980.

Question 9

Two numbers are such that the HCF of the numbers formed by squaring the sum

of the two and the difference between their squares is 10. Also, the LCM of these

new numbers is 40 times the HCF. Find these numbers.

(1) 2, 8 (2) 3, 7 (3) 5, 6 (4) 4, 7

Solution: (2)

Let the numbers be x and y. The new numbers are (x + y)2 and (x2 � y2)

The HCF is x + y and LCM is (x + y)2 (x - y)

From the question x + y = 10 and (x + y)2(x - y) z = 40 10

x = 7, y = 3.

Question 10

If in a set of three numbers, one number is also the HCF of all the three

numbers, then �

(1) One of the other two numbers will also be the LCM of all the three.

(2) For the other two numbers, one will be the HCF.

(3) The LCM of the three numbers will be the LCM of the other two.

(4) None of the above.

Solution: (3)

Let the three numbers be a, b and c.

a = HCF b = ka; c = la where k, l are integers.

LCM of (a, b and c) = LCM of (k and l) a = LCM of b and c

Solved Examples

Question 1

x and y are positive integers and 6 is a factor of xy. Which of the following must

always be true?

(1) 6 is a factor of x but only of y (2) 6 is a factor of both x and y

(3) 2 is a factor of x and y (4) 3 is a factor of at least one of x and y

Solution: (4)

An actual example will be easier to grasp the situation. If xy is 48,

48 × 1 or 16 × 3 or 12 × 4 or 24 × 2, clearly one of the two numbers in each must

have 3 as a factor. The other three choices are not applicable.

Question 2

If it were possible to mark on the number line the points 1

60and

1

12, the point

exactly midway between 1

60and

1

12will represent the number

(1) 1

20 (2)

1

(3)

24

1

30 (4)

1

40

Solution: (1)

The required number = 1 1 1 1 1 1

2 60 12 2 10 20

! "# $ % $& '

( )

Question 3

The number 88179 is not divisible by:

(1) 7 (2) 17 (3) 13 (4) 11

Solution: (4)

The divisibility rule for 11 is the easiest to try. Trying it out first, we get our

answer.

Question 4

A 2-digit number is divisible by 4 but not by 8. When the digits are interchanged,

the number is also divisible by 8. The absolute value of the difference between

such a number and its reverse is:

(1) 24 (2) 36 (3) 40 (4) 32

Solution: (2)

Both the digits have to be even. All numbers between 10 and 99 divisible by

4 but not by 8 should be odd multiples of 4. The possible numbers are 28,

44, 68 and 84. (No need to consider 20 and 60 because when the digits of these

numbers are interchanged, they form a single digit number.)

When reversed, the numbers are 82, 44, 86 and 48.

Only 48 is divisible by 8. Hence the difference is 84 – 48 = 36

Question 5

‘xyz’ denotes a 3-digit number. If ‘x’ and ‘y’ are interchanged, the value of the

number decreases by 90. How many possible values exist for ‘x’ and ‘y’?

(1) 10

(2) 9

(3) Depends on value of ‘y’

(4) None of these

Solution: (2)

Take 452 for example. If numbers are swapped it becomes 542.

Take 361 if swapped it becomes 631.

It is obvious that x, y must be consecutive numbers for difference to be 90. Since

the value of numbers decreases, y < x. Therefore possible pairs 10, 21, 32…98

which are 9 pairs.

Alternatively

100x + 10y + z –(100y + 10x + z)= 90

* 90x – 90y = 90

+x – y = 1

Also, x > 0 and x and y both are less then 10.

There are 9 such pairs of (x, y) satisfying equation (i).

Question 6

A shopkeeper is very particular that the amount for which he buys and sells

goods should always include the digit ‘9’ in it. Moreover, the digits should not

add up to 13 or a multiple of 13. If the lowest price that he can buy an item at

is Rs 400 and the highest price he can sell it for is Rs 899, the maximum profit

possible is

(1) 499

(2) 498

(3) 489

(4) 479

Solution: (4)

Take the lowest number and add 9, i.e., 400 + 9 = 409

But the digits of this number add up to 13. So the number should be 419.

Similarly, 899 adds up to 26 which is divisible by 13.

+ the number is 898.

Maximum profit = 898 – 419 = 479.

Question 7

In the following series each term after the first two is a sum of previous two

term 1, 1, 2, 3 …

What is the ratio of the number of even terms to the number of odd ones in the

first 90 terms?

(1) 2 :7 (2) 1 :3 (3) 1 : 2 (4) 1 : 1

Solution: (3)

We expand the series 1, 1, 2, 3, 5, 13, 21, 34, 55…

We find that we have two odds followed by an even, and then again two odds

and so on. First two being odds, their sum is even, which when added to its

previous odd number gives an odd number. This odd number added to its

previous even number gives another odd. This last odd number, when added to

the number before it that is odd, gives an even number and thus the cycle goes

on as OOEOOEOOE…O being odd and E being even

+EVEN : ODD = 1 : 2

Question 8

a, b, c, d and e are five consecutive given numbers. What is the maximum power

of 2 by which the product of the above numbers be necessarily divisible?

(1) 6 (2) 7 (3) 8 (4) 9

Solution: (3)

2and

2,2,2,2

edcba

are five consecutive integers. At least, one of the five is divisible by 4, one

by 2 and the rest are odd. Therefore product should be divisible by 2 to the

power of (5 + 2 + 1).

Question 9

For a 3-digit number of which one digit is 6, the sum of the number and its mirror

image is a 3-digit number divisible by 111. What is this sum?

(1) 666 (2) 777 (3) 888 (4) 999

Solution: (3)

Note that in the sum, the middle digit is added to itself and the extreme

digits are added to each other. Also note that the sum has all digits same.

Now, 6 cannot be the middle digit as 6 + 6 = 12 and 222 is not a choice.

Notice, however, that the middle digit in the sum has to be an even

number. So the viable choices are 666 and 888. But as 6 is one of the

extreme digits, the sum cannot be 666. A little inspection will reveal that the

number is 642 or 246.

Question 10

There is one number which is formed by writing one digit 6 times (e.g. 111111,

444444 etc.) Such a number is always divisible by

(1) 7, 33 and 13 (2) 7and 33 only (3)13 and 33 only (4) 7 and 13 only

Solution: (1)

A six-digit number with all identical digits, e.g., 888888 = 888 % 1001.

1001 is divisible by 7, 11 and 13. Also, the three-digit number with identical digits

will be divisible by 3. Hence, the number will be divisible by 7, 13 and 33.

Question 11

The sum of three times one natural number and twice another natural number is

less than 20. If the first natural number is less or equal to 3, find highest value of

the second natural number.

1) 9 2) 6 3) 7 4) 8

Solution: (4)

Let the first natural number be ‘x’ and the second one ‘y’.

Now 3x + 2y < 20. The first natural number can take values 1, 2 or 3 and the

second natural number would be highest only when the first natural number

takes the least value i.e. 1

If x = 1, then 3(1) + 2y < 20, Then 2y < 17. The highest value of ‘y’ would be 8.