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

S1. Ans.(a)
Sol. Let the side of the cube = a
Length of diagonal DF = AG = CE =√3 a
Circum radius of equilateral triangle having side √3 a
R=Side/√3
R=(a√3)/√3=a
Hence, radius of the circum circle is equal to the side of the cube.

S3. Ans.(b)
Sol. ∠BAC+∠ABC+∠BCA=180°  (sum of angles of a triangle)
⇒ 30° + 90° + ∠BCA=180°
⇒ ∠BCA=60°
Now, CE bisects ∠BCD (given)
∴∠ECD=30°
Now, ∠CED+∠EDC+∠DCE=180°
(sum of angles of a triangle)
∴∠CED=60°

Q4. David gets on the elevator at the 11th floor of a building and rides up at the rate of 57 floors per minute. At the same time, Albert gets on an elevator at the 51st floor of the same building and rides down at the rate of 63 floors per minute. If they continue travelling at these rates, then at which floor will their paths cross?
(a).19
(b).28
(c).30
(d).37
S4. Ans.(c)
Sol. Suppose their paths cross after x minutes.
Then, 11 + 57x = 51 – 63x
120x = 40
x =1/3
Number of floors covered by David in (1/3) min. =1/3* 57= 19.
So, their paths cross at (11 +19) i.e., 30th floor.

Q5. The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ 26.50 per metre is Rs. 5300, what is the length of the plot in metres?
(a).40
(b).50
(c).120
(d).60
S5. Ans.(d)
Sol.  Let breadth = x metres
Then, length = (x + 20) metres.
Perimeter =5300/26.5 = 200 m.
2[(x + 20) + x] = 200
2x + 20 = 100
2x = 80
x = 40.
Hence, length = x + 20 = 60 m.

Q7. The length, breadth and height of a room are in the ratio 3 : 2 : 1. If the breadth and height are halved while the length is doubled, then the total area of the four walls of the room will
(a) remain the same
(b) decrease by 13.64%
(c) decrease by 15%
(d) decrease by  30%

S7. Ans.(d)

Sol. Let the length breadth and height of the room be 3, 2 and 1 unit respectively.
Area of the four walls of the room = 2(l + b)h
= 2(3 + 2) × 1 = 10 sq unit
New length breadth and height of the room will be 6, 1 and 1/2 unit respectively
∴ New area of the four walls of the room =2(6+1)×1/2=7 sq unit
Percentage decrease =(10 – 7)/10×100%=30%

Q8. A man buys sprit at Rs 60 per litre, adds water to it and then sells it at Rs. 75 per litre. What is the ratio of spirit to water if his profit in the deal is 37.5%?
(a) 9 : 1
(b) 10 : 1
(c) 11 : 1
(d) None of these

Q9. Rs. 395 are divided among A, B and C in such a manner that B gets 25% more than A and 20% more than C. The share of A is:
(a) Rs. 198
(b) Rs. 120
(c) Rs. 180
(d) Rs. 195
S9. Ans.(b)
Sol. Suppose A get Rs. x
Then B gets 125% of
x=125/100×x=Rs.5x/4
If B gets Rs. 120
Then C gets Rs. 100
If B gets Rs. 5x/4
Then C gets
= Rs. (100/120×5x/4) = Rs. 25x/24
∴ x+5x/4+25x/24=395
⇒  (24x + 30x + 25x) = (395 × 24)
⇒ 79x = (395 × 24)
⇒ x=((395 × 24)/79)=120
Hence, A gets Rs. 120

Q10. At his usual rowing rate, Rahul can travel 12 miles downstream in a certain river in six hours less than it takes him to travel the same distance upstream. But if he could double his usual rowing rate for this 24 mile round trip, the downstream 12 miles would then take only one hour less than the upstream 12 miles. What is the speed of the current in miles per hour?
(a) 7/3
(b) 4/3
(c) 5/3
(d) 8/3
S10. Ans.(d)
Sol. Let the speed of man in still water is x miles/h and speed of the current be y miles/h
Then, 12/(x – y )-12/(x +  y)=6 …(i)
And 12/(2x – y)-12/(2x + y)=1 …(ii)
Solving equation (i) and (ii)
y=8/3

Q11. The remainder obtained when a prime number greater than 6 is divided by 6 is
(a) 1 or 3
(b) 1 or 5
(c) 3 or 5
(d) 4 or 5
S11. Ans.(b)
Sol. Let us solve the question for some prime numbers greater than 6 i.e. 7, 11, 13 and 17. If these numbers are divided by 6, the remainder is always either 1 or 5.

Q12. What is the present worth of a house which would be worth Rs 50000 after 3 years, if it depreciates at the rate of 10%
(a) Rs. 35765.74
(b) Rs. 67560.74
(c) Rs. 67655.74
(d) Rs. 68587.10
S12. Ans.(d)
Sol. Value of the house after 3 years = Rs. 50000
∴ Present worth =50000/(1-10/100)^3 =50000/(0.9)^3
= Rs. 68587.10

Q13. 5% of income of A is equal to 15% of income of B and 10% of income of B is equal to 20% of income of C. If the income of C is Rs. 2000, what is the total income of A, B and C?
(a) Rs. 18000
(b) Rs. 16000
(c) Rs. 15000
(d) None of these
S13. Ans.(a)

Sol. 10/100×B=20/100×C=B=2C

⇒ B = (2 × 2000) = 4000
5/100×A=15/100×B

⇒ A = 3B = (3 × 4000) = 12000
∴ (A + B + C) = Rs. (12000 + 4000 + 2000)
= Rs. 18000
Q14. In ∆LMN, LO is the median. Also, LO is the bisector of ∠MLN. If LO = 3 cm and LM = 5 cm, then find the area of ∆LMN
(a) 12 sq cm
(b) 10 sq cm
(c) 4 sq cm
(d) 6 sq cm

Q15. When a number is first increased by 10% and then reduced by 10% the number:
(a) Does not change
(b) Decrease by 1%
(c) Increase by 1%
(d) None of these
S15. Ans.(b)
Sol. Let the given number be x
Increased number = (110% of x)
=(110/100×x)=11x/10
Finally 11x/10, reduced number
=(90% of 11x/10)=(90/100×11x/10)=99x/100
Decrease =(x-99x/100)=x/100
Decrease % =(x/100×1/x×100 )%=1%

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