A candidate with quantitative aptitude knowledge will be in a better position to analyse and make sense of the given data. Quantitative Aptitude knowledge is an important measure for a prospective business executive’s abilities.
The article SSC CPO Quantitative Aptitude Quiz 5 provides Quantitative Aptitude questions with answers useful to the candidates preparing for Competitive exams, Entrance exams, Interviews etc. The article SSC CPO Quantitative Aptitude Quiz 5 will assist the students to know the expected questions from Quantitative Aptitude.
Answer: Option D
Explanation:
Principal
= Rs. \(\frac{100 \times 4016.25}{9 \times 5}\)
= Rs. \(\frac{401625}{45}\)
= Rs. 8925.
2. A cistern 6m long and 4 m wide contains water up to a depth of 1 m 25 cm. The total area of the wet surface is:
Answer: Option A
Explanation:
Area of the wet surface = [2(lb + bh + lh) – lb]
= 2(bh + lh) + lb
= [2 (4 x 1.25 + 6 x 1.25) + 6 x 4] m\(^{2}\)
= 49 m\(^{2}\).
3. If log\(_{10}\) 2 = 0.3010, then log\(_{2}\) 10 is equal to:
Answer: Option B
Explanation:
log\(_{2}\) 10 = \(\frac {1}{log_{10} 2}\) = \(\frac {1}{0.3010}\) = \(\frac {10000}{3010}\) = \(\frac {1000}{301}\).
4. In a 200 meters race A beats B by 35 m or 7 seconds. A’s time over the course is:
Answer: Option C
Explanation:
B runs 35 m in 7 sec.
B covers 200 m in (\(\frac {7}{35}\) x 200) = 40 sec.
B’s time over the course = 40 sec.
A’s time over the course (40 – 7) sec = 33 sec.
5. The price of 2 sarees and 4 shirts is Rs. 1600. With the same money, one can buy 1 saree and 6 shirts. If one wants to buy 12 shirts, how much shall he have to pay?
Answer: Option B
Explanation:
Let the price of a saree and a shirt be Rs. x and Rs. y respectively.
Then, 2x + 4y = 1600 …. (i)
and x + 6y = 1600 …. (ii)
Divide equation (i) by 2, we get the below equation.
=> x + 2y = 800. — (iii)
Now subtract (iii) from (ii)
x + 6y = 1600 (-)
x + 2y = 800
—————-
4y = 800
—————-
Therefore, y = 200.
Now apply the value of y in (iii)
=> x + 2 x 200 = 800
=> x + 400 = 800
Therefore x = 400
Solving (i) and (ii) we get x = 400, y = 200.
Cost of 12 shirts = Rs. (12 x 200) = Rs. 2400.
Answer: Option D
Explanation:
Let distance = x km and usual rate = y kmph.
Then, \(\frac {x}{y}\) – \(\frac {x}{y+3}\) = \(\frac {40}{60}\) => 2y(y + 3) = 9x ….(i)
And,\(\frac {x}{y-2}\) – \(\frac {x}{y}\) = \(\frac {40}{60}\) => y(y – 2) = 3x ….(ii)
On dividing (i) by (ii), we get: x = 40.
2. 18 men bind 900 books in 10 days. Find how many binders will be required to bind 600 books in 12 days?
Answer: Option A
Explanation:
We have to find the number of binders. Let the number of binders be x.
Direct Proportion:Less Books (↓),Less binders(↓)
Indirect Proportion:More days (↑),Less binders (↓)
\[ 18 : x ::
\begin{cases}
900 : 600 – – – (Books) \\
12 : 10 – – – (Days)
\end{cases}
\]
x × 900 × 12 = 18 × 600 × 10
x = \(\frac {18 \times 600 \times 10}{900 \times 12}\) = 10
3. Find the value of \(^{20}c_{17}\)
Answer: Option B
Explanation:
\(^nc_{r}\) =\(\frac {^np_{r!}}{r!}\)
\(^nc_{r}\) = \(\frac {n!}{(r!) (n – r)!}\)
\(^{20}c_{17}\) =\(\frac {20!}{(17!) (20 – 17)!}\)
\(^{20}c_{17}\) =\(\frac {20 \times 19 \times 18 \times 17!}{(17!) (3)!}\)
\(^{20}c_{17}\) =\(\frac {20 \times 19 \times 18}{3 \times 2 \times 1}\)
\(^{20}c_{17}\) = 1140.
4. (1331)\(^{- (2/3)}\)
Answer: Option C
Explanation:
Cube root of 1331 is 11. Therefore,
(11)\(^{3 \times -(2/3)}\)
Remember the law of indices (xm)n = xmn
(11)\(^{3 \times -(2/3)}\) = 11\(^{-2}\)
x\(^{-1}\) = \(\frac {1}{x}\)
Hence, 11\(^{-2}\) = \(\frac {1}{112}\) = \(\frac {1}{112}\)
5. Two pipes can fill a tank in 8 hrs & 6 hrs respectively. If they are opened on alternate hours and if pipe A gets opened first, then in how many hours, the tank will be full?
Answer: Option B
Explanation:
Pipe A’s work in 1 hr = \(\frac {1}{8}\)
Pipe B’s work in 1 hr = \(\frac {1}{6}\)
Pipes (A+B)’s work in first 2 hrs when they are opened alternately = 1/8 + 1/6 = 7 /24
Now,
In 4 hrs they fill : 2 X (\(\frac {7}{24}\)) = \(\frac {7}{12}\)
In 6 hrs they fill : 3 X (\(\frac {7}{24}\)) = \(\frac {7}{8}\)
After 6 hrs, part left empty = \(\frac {1}{8}\)
Now it is A’s turn to open up.
In one hr it fills \(\frac {1}{8}\) of the tank.
So, the tank will be full in = 6 hrs + 1 hr = 7 hrs.
Answer: Option C
Explanation:
log (2\(^{64}\)) = 64 x log 2
= (64 x 0.30103)
= 19.26592
Its characteristic is 19.
Hence, then number of digits in 2\(^{64}\)6 is 20.
2. In a race of 800m, if the ratio of the speeds of two participants P and R is 3:4, and P has a start of 120m, then R beats P by:
Answer: Option B
Explanation:
As P has a start of 120m, P has to cover 680m while R has to cover 800m.
Now, when P covers 3m, R covers 4m.
Hence, when R covers 800m, P covers (800 * 3) / 4 = 600m
Therefore, R beats P by 80m.
3. In a camp, there is a meal for 120 men or 200 children. If 150 children have taken the meal, how many men will be catered to with remaining meal?
Answer: Option B
Explanation:
There is a meal for 200 children. 150 children have taken the meal.
The remaining meal is to be catered to 50 children.
Now, 200 children = 120 men.
50 children = \(\frac {120 \times 50}{2500}\) = 30 men.
4. Using all the letters of the word “NOKIA”, how many words can be formed, which begin with N and end with A?
Answer: Option B
Explanation:
There are five letters in the given word.
Consider 5 blanks
The first blank and last blank must be filled with N and A all the remaining three blanks can be filled with the remaining 3 letters in 3! ways.
The number of words = 3! = 6.
4. (\(\frac {x^{b}}{x^{c}})^{b + c – a}\) . (\(\frac {x^{c}}{x^{a}})^{c + a – b}\) . (\(\frac {x^{a}}{x^{b}})^{a + b – c}\) = ?
Answer: Option B
Explanation:
Given Exp.
= x\(^{(b – c)(b + c – a)}\) . x\(^{(c – a)(c + a – b)}\) . x\(^{(a – b)(a + b – c)}\)
= x\(^{(b – c)(b + c) – a(b – c)}\) . x\(^{(c – a)(c + a) – b(c – a)}\)
. x\(^{(a – b)(a + b) – c(a – b)}\)
= x\(^{(b^{2} – c^{2} + c^{2} – a^{2} + a^{2} – b^{2})}\) . x\(^{-a(b – c) – b(c – a) – c(a – b)}\)
= (\(x^{0} \times x^{0}\))
= (1 x 1) = 1.