Quantitative Aptitude Practice Test for SSC CGL (TIER-I)

QUANTITATIVE APTITUDE PRACTICE TEST

1.In a three digit number, the digit at the hundred's place is two times the digit at the unit's place and the sum of the digits is 18. If the digits are reversed, the number is reduced by 396. The difference of hundred's and ten's digit of the number is:
a)2
b)5
c)1
d)3


2.Which of the following numbers will always divide a six digit number of the form xyxyxy (where 1 ≤ x ≤ 9, 1 ≤ y ≤ 9)?
a)11011
b)11010
c)10101
d)1010

3.The H.C.F of two numbers is 96 and their L.C.M is 1296. If one of the numbers is 864, the other is:
a)135
b)144
c)132
d)140

4.If the sum of the interior angles of a regular polygon be 1080°, the number of sides of the polygon is :
a)9
b)5
c)7
d)8

5.The diameter of a cylinder is 7cm and its height is 16cm. The lateral surface area of the cylinder is:
a)340 sq.cm
b)352 sq.cm
c)335 sq.cm
d)360 sq.cm

6.A man and a woman working together can do a certain work in 18 days. Their skills in doing the work are in the ratio 3:2. How many days will the woman take to finish the work alone?
a)40
b)30
c)20
d)45

7.The single discount which is equivalent to successive discounts of 20%, 15% and 10% is :
a)38.8%
b)33.35%
c)29.56%
d)42.55%

8.An article of CP Rs.8000 is marked at Rs.11200. After allowing a discount of x% a profit of 12% is made. What is the value of x?
a)20%
b)25%
c)18%
d)12%

9.75gm of sugar solution has 30% sugar in it. Then the quantity of sugar that should be added to the solution to make the quantity of the sugar 70% in the solution is :
a)160 gm
b)140 gm
c)100 gm
d)110 gm

10.If a boat goes 100 km down-stream in 10 hours and 75 km upstream in 15 hours, then the speed of the stream is:
a)2.5 km/hr
b)4 km/hr
c)5 km/hr
d)5.5 km/hr

Answer key

1.a)
2.c)
3.b)
4.d)
5.b)
6.d)
7.a)
8.a)
9.c)
10.a)

Solution

1.Let, the number be 100(2x) + 10y + x = 201x + 10y.....(i)
now, 2x + y + x = 18
or, 3x + y = 18.....(ii)
When the digits are reversed, number = 100(x) + 10y + 2x = 102x + 10y....(iii)
so, 201x + 10y - 102x - 10y = 396
                 or, x = 4
From eqn (i)
3 × 4 + y = 18
or, y = 6
Required difference = 2x - y = 2 × 4 - 6 = 2

2.Number = xyxyxy
= xy × 10000 + xy × 100 + xy
= xy (10000 + 100 + 1)
= xy × 10101

3.First number × second number = HCF × LCM
or, 864 × second number = 96 × 1296
or, second number = (96 × 1296)/864 = 144

4.Sum of the interior angles of a regular polygon of 'n' sides = (2n - 4) × 90°
Therefore, (2n - 4) × 90° = 1080°
                     or, n = 8

5.Lateral surface area of the cylinder = 2πrh = 2 × 22/7 × 7/2 × 16 = 352 sq.cm

6.Man : Woman (efficiency) = 3 : 2
i.e, woman completes 2/5th work in 18 days.
Therefore, time taken by the woman to complete the whole work = (18 × 5)/2 = 45 days

7.Single equivalent discount for 20% and 15% = [20 + 15 - {(20 × 15)/100}] = 32%
single equivalent discount for 32% and 10% = [32 + 10 - {(32 × 10)/100}] = 38.8%

8.SP for a profit of 12% = (8000 × 112)/100 = Rs.8960
Therefore, discount = 11200 - 8960 = Rs.2240
If the discount percent be x,
then (11200 × x)/100 = 2240 or, x = 20%

9.Sugar in original solution = (75 × 30)/100 = 22.5gm
Let, xgm of sugar be mixed.
so, (22.5 + x)/(75 + x) × 100 = 70
     or, x = 3000/30 = 100gm

10.Rate of downstream = 10 kmph
Rate of upstream = 5 kmph
so, speed of the current = 1/2 (10 - 5) = 2.5 kmph