# Special Quant Quiz (Arithmetic) SSC CGL Tier-II 2016

Q1. The sum of the digits of a two-digit number is 1/11 of the sum of the number and the number obtained by interchanging its digits. What is the difference between the digits of the number?
(a) 3
(b) 0
(c) 1
(d) Data is insufficient
S1. Ans.(d)
Sol. x + y = 1/11 (10x + y + 10y + x)
x + y = 1/11 (11x + y) …(i)
From equation (i) we can’t determine difference between the digit of the number.

Q2. 2! + 4! + 6! + 8! + 10! … 100! When divided by 5, would leave remainder
(a) 0
(b) 1
(c) 2
(d) 3
S2. Ans.(b)
Sol. 6! + 8! + 8! … 100! always divisible by 5.  because there is a common factor 5 in every term.
So required answer = (2! + 4!)/5=26/5
Remainder = 1

Q4. In how many ways can a number 6084 be written as a product of two different factors?
(a) 27
(b) 26
(c) 13
(d) 14

Q5. P + Q + R + S is odd. At least how many of these (is/are) odd?
(a) 0
(b) 1
(c) 2
(d) Cannot be determine
S5. Ans.(b)
Sol. If 1 odd + 3 even number them sum will be odd.
So there should be at least number one number will be odd.

Q6. When a two-digit number is divided by the sum of its digits, the quotient is 4. If the digits are reversed, the new number is 6 less than twice the original. The number is
(a) 12
(b) 21
(c) 42
(d) 24
S6. Ans.(d)
Sol. Let the two digit number is xy
(10x+y)/(x+y)=4
10x + y = 4(x+ y)
6x = 3y …(i)
Again, now,
10y + x = 2(10x + y) – 6
10y + x = 20x + 2y – 6
19x – 8y = 6 …(ii)
By solving these two equation,
x = 2
y = 4
Number will be 24.

Q7. How many numbers between 200 and 600 are divisible by 4, 5 and 6?
(a) 5
(b) 6
(c) 7
(d) 8
S7. Ans.(b)
Sol. If number is divisible by (4, 5 & 6), then number must be divisible by  LCM of (4, 5, 6) = 60
So first number which is divisible= 240
Last number = 540
Total number divisible by 60 = (540-240)/60+1
= 5 + 1 = 6

S8. Ans.(a)
Sol. a. Since the first digit of (a 3 b) is written as it is after subtracting ac from it. It means that there is no carry over from a to 3.
b. There must be a carry over from 3 to b, because if no carry over is there, it means 3 – a = a.
⇒ 2a = 3 ⇒ a = 3/2
Which is not possible because a is a digit. For a carry over 1, 2 – a = a
⇒ a = 1
c. Now 10 + b – c = 9
⇒ b – c = –1
It means b and c are consecutive digits (2, 3), (3, 4), …, (8, 9).
So, there are 7 pairs

Q9. Find the value of 896 × 896 – 204 × 204.
(a) 781200
(b) 761200
(c) 771200
(d) 764200
S9. Ans.(b)
Sol. a^2-b^2 = (a + b) (a – b)
(where a = 896 and b = 204)
= (896 + 204) (896 – 204)
= 1100 × 692 = 761200
Q10. Ajay went on a ten-mile drive of his new imported bike. He started with a certain speed and after covering each mile, his speed is decreased by 20% for the next mile. If he took 5 minutes to cover the first five mile of the drive, what is the approximate time taken by him to cover the next five miles?
(a) 14 minutes and 1 second
(b) 15 minutes and 15 seconds
(c) 16 minutes and 16 seconds
(d) 17 minutes and 17 seconds

Q11. The metro service has a train going from Mumbai to Pune and Pune to Mumbai every hour the first one at 6 a.m. The trip from one city to other takes 4.5 hours, and all trains travel at the same speed. How many trains will you pass while going from Mumbai to Pune if you start at 12 noon?
(a) 8
(b) 10
(c) 9
(d) 13

Q12. A tiger is 50 of its own leaps behind a deer. The tiger takes 5 leaps per minute to the deer’s 4. If the tiger and the deer cover 8m and 5m per leap respectively, what distance will the tiger have to run before it catches the deer?
(a) 600 m
(b) 700 m
(c) 800 m
(d) 1000 m

Q13. A man travelled a distance of 61 km in 9 hours. partly on foot at the rate of 4 km/hr and partly on bicycle at the rate of 9 km/hr. The distance travelled on foot was:
(a) 12 km
(b) 16 km
(c) 20 km
(d) 24 km
S13. Ans.(b)
Sol. Let man walked for t hours.
∴ t × 4 + (9 – t) × 9 = 61
⇒ 4t + 81 – 9t = 61
⇒ 5t = 20 ⇒ t = 4 hours
∴ Distance travelled on foot = 4 × 4 = 16 km.

Q14. A car travels from P to Q at a constant speed. If its speed were increased by 10/h, it would have taken one hour lesser to cover the distance. It would have taken further 45 min lesser if the speed was further increased by 10 km/h. What is the distance between two cities?
(a) 420 km
(b) 540 km
(c) 600 km
(d) 650 km
S14. Ans.(a)
Sol. Assume speed = v km/h, distance = d km
From question: d/v-d/((v+10))=1
d[v + 10 – v] = v(v + 10)
10d = v(v + 10) …(i)
From IInd condition:
d/v-d/((v+20))=1+45/60=7/4
(d[v+20-v])/(v(v+20))=7/4
80d = 7v(v + 20) …(ii)
From (i) & (ii)
80d = 7v[(v + 10) + 10]
80d = 7[v(v + 10) + 10v]
80d = 7[10d + 10v]
80d = 70d + 70v
10d = 70v
d = 7v
v = d/7 put in equation (i)
10d = d/7 [d/7+10]
10=d/49+10/7
d/49=10-10/7=60/7
d/49=60/7 ⇒ d = 420 km.

Q15. A dog after travelling 50 km meets a swami who counsels him to go slower. He then proceeds at 3/4 of his former speed and arrives at his destination 35 minutes late. Had the meeting occurred 24 km further the dog would have reached its destination 25 minutes late. The speed of the dog is:
(a) 48 km/h
(b) 36 km/h
(c) 54 km/h
(d) 58 km/h
S15. Ans.(a)
Sol. For 24 km. when dog reduced his speed to 3/4 th if the former speed dog take time (35 – 25) = 10 minutes.
Let the starting speed of dog n km/hr and it takes t time to travel 24 km.
According to the question,
(n/3n)/4=(t+1/6)hr/thr
t = 30 minute

Speed of dog = 24/(1/2 hr) = 48 km/hr

×
OR

×