UPSC Question-Paper Contributed by sanalKumar updated on Oct 2020
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UPSC IES/ISS General ability, Aptitude questions with answers and explanation, UPSC previous years solved questions with answers
Part - C UPSC questions with answers and explanations
Numerical Aptitude


101. If p = 124, 3√p (p2 + 3p + 3) + 1 = ?a. 5
b. 7
c. 123
d. 125 (Ans)
Ans :
3√p (p2 + 3p + 3) + 1
= 3√p3 + 3p2 + 3p + 1
= 3√(p + 1)3= (p + 1)
= 125

102. If √1 - x3/100 = 3/5, then x equals —
a. 2
b. 4 (Ans)
c. 16
d. (136)1/3

Ans : √1 - x3/100 = 3/5⇒ 1 - x3/100 = 9/25∴ x3/100 = 16/25⇒ x3 = 16*100/25 = (4)3∴ x = 4 103. I multiplied a natural number by 18 and another by 21 and added the products. Which one of the following could be the sum?a. 2007 (Ans)
b. 2008
c. 2006
d. 2002
Ans : 18x + 21y = 3 (6x + 7y)Only 2007 is divisible by 3. Hence it is only possible.

104. The product of two numbers is 45 and their difference is 4. The sum of squares of the two numbers is —a. 135
b. 240
c. 73
d. 106 (Ans)Ans : x y = 45 x - y = 4∴ x2 + y2 = (x - y)2 + 2xy = 16 + 90 = 106


105. √8 + √57 + √38 + √108 + √169 = ?a. 4 (Ans)
b. 6
c. 8
d. 10
Ans : ? = √8 + √57 + √ 38 +√108 + √ 169

= √ 8 + √ 57 + √38 + √108 +13 
= √ 8 + √57 + 7 = √ 8 + 8 = 4

106. The square root of 14 + 6√5 is —
a. 2 + √5
b. 3 + √5 (Ans)
c. 5 + √3
d. 3 + 2√5
Ans : √ 14 + 6 √ 5 = √9 + 5 + 2*3* √5= 3 + √5

107. When 231 is divided by 5 the remainder is —
a. 4
b. 3 (Ans)
c. 2
d. 1
Ans : The unit digit in (2)31 will be same as the unit digit in (2)3, because the last digit is repeated after each 4 index∴ (2)31 → 228 + 3 = 23 = 8∴ After dividing by 5, the remainder will be = 8 - 5 = 3.
108. The value of 1 + 1 is 1 + 1 1 + 1 1 + 1 1 + 2/3a. 21/13
b. 17/3
c. 34/21 (Ans)
d. 8/5
Ans : 1 + 1 is 1 + 1 1 + 1 1 + 1 1 + 2 /3

= 1 + 1 1 + 1 1 + 1 1 + 3/5 
= 1 + 1 1 + 1 1 + 5/8 
= 1 + 1 1 + 8/13 = 1 + 13/21 = 34/21109. The unit digit in the product (122)173 is —
a. 2 (Ans)
b. 4
c. 6
d. 8
Ans : A

110. The value of 2 + √3 / 2 - √3 + 2 - √3 / 2 + √3 + √3 + 1 / √3 - 1 is—a. 16 + √3 (Ans)b. 4 - √3c. 2 - √3d. 2 + √3
Ans : 2 + √ 3 / 2 - √3 + 2 - √3 / 2 + √3 + √3 + 1 / √3 - 1= (2 + √3)2 + (2 - √3)2 / ( 2 - √3) + (2 + √3) + (√3 + 1) (√3 + 1) / (√3 - 1) (√3 + 1)= 14/1 + 3 + 1 + 2√3/2
= 14 + 2 + √3
= 16 + √3
111. If a*b = 2a + 3b - ab, then the value of (3*5 + 5*3) is—a. 10 (Ans)
b. 6
c. 4
d. 2
Ans : a * b = 2a + 3b - ab∴ (3 * 5 + 5 * 3) = (2 * 3 + 3 * 5 - 3 * 5) + (2 * 5 + 3 * 3 - 5 * 3)= (6 + 15 - 15) + (10 + 9 - 15)
= 6 + 4
= 10

112. Simplify : 0.0347 * 0.0347 * 0.0347 + (0.9653)3 / (0.0347)2 - (0.347) (0.09653) + (0.9653)2a. 0.9306
b. 1.0009
c. 1.0050
d. 1 (Ans) (Ans)
Ans : 0.0347 * 0.0347 * 0.0347 + (0.9653)3 / (0.0347)2 - (0.347) (0.09653) + (0.9653)2= (0.0347)3 + (0.9653)3 / (0.0347)2 - (0.0347) (0.9653) + (0.9653)2 (0.0347 + 0. 9653)= [(0.0347)2 - (0.0347 * 0.9653) + (0.9653)2] / [(0.0347)2 - 0.0347 * 0.9653 + (0.9653)2]
= 1

113. A copper wire is bent in the form of an equilateral triangle and has area 121 √3 cm2, If the same wire is bent into the form of a circle, the area (in cm2) enclosed by the wire is ( Take = 22/7)—
a. 364.5
b. 693.5
c. 346.5 (Ans)
d. 639.5
Ans : Area of equilateral ? = √3/ 4 (side)2∴ 121 √3 = √3/4 (side)2∴ (side) = √4 *121 = 22 cm∴ Length of wire = 3 * 22 = 66cm∴ 2 * 22/7 * r = 66r = 66 * 7 / 2 * 22
= 21 / 2 cm
∴ Area of the circle = 22/7 * 21/2 * 21/2= 346.5 cm2

114. A child reshapes a cone made up of clay of height 24 cm and radius 6 cm into a sphere. The radius (in cm) of the sphere is—a. 6 (Ans)
b. 12
c. 24
d. 48
Ans : 3/4 r3 = 1/3 (6)2 * 24∴ r3 = 216 = (6)3∴ r = 6 cm

115. Water flows into a tank which is 200 m long and 150 m wide, through a pipe of crosssection 0.3 m * 0.2 m at 20 km/hour. Then the time (in hours) for the water level in the tank to reach 8 m is—
a. 50
b. 120
c. 150
d. 200 (Ans)
Ans : Reqd. time = 200 * 150 * 8/ 0.3 * 0.2 * 20000 hrs.= 200 hrs.

116. The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is—
a. 1
b. 2 (Ans)
c. 3
d. 4
Ans : Let the numbers be 13x and 13y.∴ 13x * 13y = 2028
∴ xy = 12∴ x = 3 and y = 4or x = 1 and y = 12∴ The number of pairs are = 2

117. Two equal vessels are filled with the mixtures of water and milk in the ratio of 3 : 4 and 5 : 3 respectively. If the mixtures are poured into a third vessel, the ratio of water and milk in the third vessel will be—a. 15 : 12b. 53 : 59c. 20 : 9d. 59 : 53 (Ans)Ans : Reqd. ratio = (3/7 + 5/8) / (4/7 + 3/8)= (24 + 35)/ (32 + 21)
= 59 : 53
118. I am three times as old as my son. 15 years hence, I will be twice as old as my son. The sum of our ages is—
a. 48 years
b. 60 years (Ans)
c. 64 years
d. 72 years
Ans : Let the present age of the son be x years∴ My present age = 3x years∴ 3x + 15 / x + 15 = 2/1
∴ Reqd. sum = 45 + 15 = 60 years

119. Three bells ring simultaneously at 11 a.m. They ring at regular intervals of 20 minutes, 30 minutes, 40 minutes respectively. The time when all the three ring together next is—a. 2 p.m.
b. 1 p.m. (Ans)
c. 1.15 p.m.
d. 1.30 p.m.
Ans : L.C.M. of 20, 30 and 40 minutes= 120 minutes= 2 hours.
∴ Reqd. time = 1 p.m.

120. A and B together can do a work in 12 days. B and C together do it in 15 days, If A's efficiency is twice that of C, then the days required for B alone of finish the work is—
a. 60
b. 30
c. 20 (Ans)
d. 15
Ans : Let the time taken by A to finish the work be x days∴ C will take 2x days to finish the work
∴ Work of B for 1 day = 1/12 - 1/x
and work of B for 1 day = 1/15 - 1/2x∴ 1/12 -1/x = 1/15 - 1/2x
∴ 1/2x = 1/12 - 1/15
= 1/60
∴ x = 30
∴ Work of B for 1 day = 1/12 - 1/30= 1/20
∴ B alone will finsh the work in 20 days. 121. A and B can do a work in 12 days, B and C can do the same work in 15 days, C and A can do the same work in 20 days. The time taken by A, B and C to do the same work is—a. 5 days
b. 10 days (Ans)
c. 15 days
d. 20 days
Ans : Reqd. time= 12 * 15 * 20 * 2/ (12 * 15 + 15 * 20 + 12 * 20)
= 7200/(180 + 300 + 240)
= 10 days

122. A is 50% as efficient as B . C does half of the work done by A and B together. If C alone does the work in 20 days, then A, B and C together can do the work in—a. 5 2/3 days
b. 6 2/3 days (Ans)
c. 6 days
d. 7 days
Ans : Let the time taken by B to complete the work be x days∴ Time ,, ,, by A ,, ,,= 2x days
∴ Work done by by C for 1 day
= 1/2 (1/x + 1/2x)
= 3/4x
∴ 3/4x = 1/20⇒ x = 15∴ Work done by (A + B + C) for 1 day = 1/30 + 1/15 + 1/20 = 9/60= 3/20
∴ Reqd. time = 6 2/3 days

123. The ratio of the volumes of water and glycerine in 240 cc of mixture is 1 : 3. The quantity of water (in cc) that should be added to the mixture so that the new ratio of added to the mixture so that the new ratio of the volumes of water and glycerine becomes 2 :3 is—a. 55
b. 60 (Ans)
c. 62.5
d. 64
Ans : The quantity of water to be added= 240 (3-2)/ (1 + 3)
= 60 c.c.

124. At present, the ratio of the ages of Maya and Chhaya is 6 : 5 and fifteen years from now, the ratio will get changed to 9 : 8 Maya's present age is—a. 21 years
b. 24 years
c. 30 years (Ans)
d. 40 years
Ans : Let the present ages of Maya and Chhaya be 6x and 5x years respectively.6x + 15/ 5x + 15 = 9/8
48x + 120 = 45x + 135
x = 5
∴ Present age of Maya is = 30 years.

125. The ratio of the income to the expenditure of a family is 10 : 7. If the family's expenses are Rs.10,500, then savings of the family is—a. Rs.4,500 (Ans)
b. Rs.10,000
c. Rs.4,000
d. Rs.5,000
Ans : Let the income and expenditures be Rs.10x and Rs.7x respectively∴ 7x = 10500⇒ x = Rs.1500∴ Savings = (10-7) * 1500= Rs.4500

126. The average mathematics marks of two Sections A and B of Class IX in the annual examination is 74. The average marks of Section A is 77.5 and that of Section B is 70. The ratio of the number of students of Section A and B is—a. 7 : 8
b. 7 : 5
c. 8 : 7 (Ans)
d. 8 : 5
Ans : Let the numbers of A and B be n and P respectively∴ n * 77.5 + P * 70 = (n + P) * 7477.5n - 74n = 74P - 70P
3.5n = 4P
∴ n/P = 4 : 3.5= 8 : 7

127. The average weight of a group of 20 boys was calculated to be 89.4 kg and it was later discovered that one weight was misread as 78 kg instead of 87 kg. The correct average weight is—a. 88.95 kg
b. 89.25 kg
c. 89.55 kg
d. 89.85 kg (Ans)
Ans : The correct average wt.
= 20 * 89.4 + 87 - 78/ 20
= 1788 +9/20
= 89.85 kg.

128. The diameter of a wheel is 98 cm. The number of revolutions in which it will have to cover a distance of 1540 m is—
a. 500 (Ans)
b. 600
c. 700
d. 800
Ans : No. of revolutions= 1540 * 100 * 7/98 * 22
= 500

129. In an equilateral triangle ABC of side 10cm, the side BC is trisected at D. Then the length (in cm) of AD is—
a. 3√7
b. 7√3
c. 10√7/3 (Ans)
d. 7√10/3Ans : cos 60 = (10)2 + (10/3)2 - AD2/ 2 * 10 * 10/3 
1/2 = 100 + 100/9 - AD2 / 200/31/2 * 200/3 = 1000 - 9 AD2 /9
AD2 = 700/9
∴ AD = 10√7/3 cm 



130. The cost price of an article is Rs.800. After allowing a discount of 10% , a gain of 12.5% was made. Then the marked price of the article is—a. Rs.1,000 (Ans)
b. Rs.1,100
c. Rs.1,200
d. Rs.1,300
Ans : S. P. of the article = 800 * 112.5/100= Rs.900
∴ M.P. of the article = 100 * 900/ (100 - 10)
= Rs.1000

131. A man bought an article listed at Rs.1,500 with a discount of 20% offered on the list price. What additional discount must be offered to the man to bring the net price to Rs.1,104 ?a. 8% (Ans)
b. 10%
c. 12%d. 15%
Ans : S. P. the article after a discount of 20%
= 1500 * 80/100
= Rs.1200
∴ Additional discount = (1200 - 1104) * 100/1200= 8%

132. If a/b = c/d = e/f = 3, then 2a2 + 3c2 + 4e2/ 2b2 + 3d2 + 4f2 = ?a. 2
b. 3
c. 4
d. 9 (Ans)
Ans : a/b = c/d = e/f = 3∴ a = 3bc = 3d
and e = 3f∴ ? = 2a2 + 3c2 + 4e2/ 2b2 +3d2 + 4f2= 2 * 9b2 + 3 * 9d2 + 4 * 9f2/ 2b2 + 3d2 + 4f2
= 9

133. The floor of a room is of size 4 m * 3m and its height is 3 m. The walls and ceiling of the room require painting. The area to be painted is—
a. 66 m2
b. 54 m2 (Ans)
c. 43 m2
d. 33 m2
Ans : Reqd. area = 2(4 + 3) * 3 + 4 * 3
= 42 + 12
= 54 m2

134. When the price of an article was reduced by 20%, its sale increased by 80%. What was the net effect on the sale?
a. 44% increase (Ans)
b. 44% decrease
c. 66% increase
d. 75% increase
Ans : Reqd. % effect = [80 - 20 - 80*20/100]%= 44% increase

135. The price of sugar goes up by 20%. If a housewife wants the expenses on sugar to remain the same, she should reduce the consumption by—
a. 15 1/5%b. 16 2/3% (Ans)c. 20%
d. 25%
Ans : Reqd. % reduction =20 * 100/ (100 + 20)%
= 16 2/3%

136. In a factory 60% of the workers are above 30 years and of these 75% are males and the rest are females. If there are 1350 male workers above 30 years, the total number of workers in the factory is—a. 3000 (Ans)
b. 2000
c. 1800
d. 1500
Ans : Let the total no.of workers be x.
∴ No. of males above 30 years
⇒ x * 60/100 * 75/100 = 1350
∴ x = 1350 * 100 * 100/ 60 * 75= 3000

137. Walking at 3/4 of his usual speed, a man is 1 1/2 hours late. His usual time to cover the same distance, in hours is—a. 4 1/2 (Ans)
b. 4
c. 5 1/2
d. 5
Ans : Let the usual time be x hours x * usual speed= 3/4 * usual speed * (x + 3/2)
x = 3/4 x + 9/8
∴ x = 9/8 * 4= 4 1/2 hours

138. The selling price of 10 oranges is the cost price of 13 oranges. Then the profit percentages is—a. 30% (Ans)
b.10%
c. 13%
d. 3%
Ans : Profit % = (13-10)/10 * 100%
= 30%

139. The marked price of a radio is Rs.480. The shopkeeper allows a discount of 10% and gains 8%. If no discount is allowed, his gain percent would be—a. 18%
b. 18.5%
c. 20.5%
d. 20% (Ans)
Ans : C.P. of the radio = 480 * 90/ 100 * 100/108= Rs.400
∴ New profit % = 480 - 400/400 * 100%=20%

140. A man sold 20 apples for Rs.100 and gained 20%. How many apples did he buy for Rs.100?a. 20
b. 22
c. 24 (Ans)
d. 25
Ans : ? C.P. of 1 apple = Rs. 100/20 * 100/120∴ No. of reqd. apples = 100 * 20 * 120/ 100 * 100= 24

141. A rectangular sheet of metal is 40 cm by 15 cm. Equal squares of side 4 cm are cut off at the corners and the remainder is folded up to form an open rectangular box. The volume of the box is—a. 896 cm3 (Ans)
b. 986 cm3
c. 600 cm3
d. 916 cm3
Ans : Vol. of the open box
= (40 - 8) * (15 - 8) * 4
= 32 * 7 * 4
= 896 cm3
142. If 78 is divided into three parts which are in the ratio 1 : 1/3 : 1/6, the middle part is—a. 9 1/3
b. 13
c. 17 1/3 (Ans)
d. 18 1/3
Ans : Ratio = 1 : 1/3 : 1/6= 6 : 2 : 1
∴ The middle part = 2 * 78/(6 + 2 + 1)= 52/3
= 17 1/3

143. The simple interest on a sum of money is 1/9 of the principal and the number of years is equal to rate per cent per annum. The rate per annum is—a. 3%
b. 1/3%
c. 3 1/3% (Ans)
d. 3/10%
Ans : ? 1/9 * P = P * R * R/100∴ R = 10/3%= 3 1/3%

144. The difference between simple interest and compound interest of a certain sum of money at 20% per annum for 2 years is Rs.48. Then the sum is—a. Rs.1,000
b. Rs.1,200 (Ans)
c. Rs.1,500
d. Rs.2,000
Ans : ? D = P (r/100)2⇒ 48 = P (20/100)2
∴ P = 48 * 25= Rs.1200

145. Shri X goes to his office by scooter at a speed of 30 km/h and reaches 6 minutes earlier. If he goes at a speed of 24 km/h, he reaches 5 minutes late. The distance to his office is—a. 20 km
b. 21 km
c. 22 km (Ans)
d. 24 km
Ans : Reqd. distance = 30 * 24/(30 - 24) * (6 + 5)/60= 720 * 11/ 6 * 60
= 22 km

146. A sum of money becomes eight times in 3 years, if the rate is compounded annually. In how much time will the same amount at the same compound rate become sixteen times?a. 6 years
b. 4 years (Ans)
c. 8 years
d. 5 years
Ans : Reqd. time = a log q/ log p= 3 * log 16/ log 8
= 3 * 4 log 2/ 3 log 2
= 4 years

Directions - The pie chart given below shows the spending of a family on various heads during a month. Study the graph and answer questions 147 to 150.(Image)


147. If the total income of the family is Rs.25,000, then the amount spent on Rent and Food together is—a. Rs.17,250
b. Rs.14,750 (Ans)
c. Rs.11,250
d. Rs.8,500
Ans : Reqd. expenditure = (45 + 14) * 25000/100
= Rs.14750

148. What is the ratio of the expenses on Education to the expenses on Food?
a. 1 : 3 (Ans)
b. 3 : 1
c. 3 : 5
d. 5 : 3
Ans : Reqd. ratio = 15 : 45
= 1 : 3

149. Expenditure on Rent is what per cent of expenditure on Fuel ?
a. 135%
b. 156% (Ans)
c. 167%
d. 172%
Ans : Reqd. ratio % = 14 * 100/9%= 156%
150. Which three expenditures together have a central angle of 1080 ?a. Fuel, Clothing and Others
b. Fuel, Education and Others (Ans)
c. Clothing, Rent and Others
d. Education, Rent and others
 

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