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Aptitude General Aptitude Test Practice Q&A - Easy Page: 4
2562.$\dfrac{4.2 \times 4.2 - 1.9 \times 1.9}{2.3 \times 6.1}$ is equal to :
0.5
1.0
20
22
Explanation:

Given Expression =$\dfrac{(a^2 - b^2)}{(a+b)(a-b)}$=$\dfrac{(a^2 - b^2)}{(a^2 - b^2)}$=1

2574.$\dfrac{5\times1.6 - 2 \times 1.4}{1.3}$=?
0.4
1.2
1.4
4
Explanation:

Given Expression =$ \dfrac{8 - 2.8}{1.3} $=$ \dfrac{5.2}{1.3} $=$ \dfrac{52}{13} $= 4.

2810.Two pipes A and B can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?
10 min. 20 sec.
11 min. 45 sec.
12 min. 30 sec.
14 min. 40 sec.
Explanation:

Part filled in 4 minutes = 4$ \left(\dfrac{1}{15} +\dfrac{1}{20} \right) $=$ \dfrac{7}{15} $.

Remaining part =$ \left(1 -\dfrac{7}{15} \right) $=$ \dfrac{8}{15} $.

Part filled by B in 1 minute =$ \dfrac{1}{20} $

$\therefore \dfrac{1}{20} $:$ \dfrac{8}{15} $:: 1 : $ x $

$ x $ =$ \left(\dfrac{8}{15} \times 1 \times 20\right) $= 10$ \dfrac{2}{3} $min = 10 min. 40 sec.

$\therefore$ The tank will be full in $\left(4 min. + 10 min. + 40 sec.\right)$ = 14 min. 40 sec.

2823.Two pipes A and B can fill a cistern in 37$ \dfrac{1}{2} $ minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the B is turned off after:
5 min.
9 min.
10 min.
15 min.
Explanation:

Let B be turned off after $ x $ minutes. Then,

Part filled by $\left(A + B\right)$ in $ x $ min. + Part filled by A in $\left(30 - x \right)$ min. = 1.

$\therefore x \left(\dfrac{2}{75} +\dfrac{1}{45} \right) $+ $\left(30 - x \right)$.$ \dfrac{2}{75} $= 1

$\Rightarrow \dfrac{11x}{225} $+$ \dfrac{(60 -2x)}{75} $= 1

$\Rightarrow$ 11$ x $ + 180 - 6$ x $ = 225.

$\Rightarrow x $ = 9.

44339.A 100 m long 3 m high and 30 cm wide wall is built by 30 men, 20 women and 50 children working 9 hours a day in 20 days. How long a wall 1.5 m high 30 cm wide can be built by 15 men, 25 women and 35 children working 2 hour a day in 15 days (given men, women and children are equally efficient)?
75 m
25 m
50 m
100 m
Explanation:

Earlier dimensions of the wall = 100 × 3 × 0.30.
New dimensions = L × 1.5 × 0.3.
∴ As men, women and children are given to be equally efficient, so in the first case, the total number of persons is 100 (i.e. 30 + 20 + 50)
and the same in the second case is 75 (15 + 25 + 35).
Length of wall = L = (75x100) x (2x9) x (15x20) x (100 x 3 x 3 x 0.3)/(1.5 x 0.3) ⇒ L = 25 m.

44366.In how many different ways can the letters of the word 'RUMOUR' be arranged?
128
130
180
200
Explanation:

The word 'RUMOUR' has 6 letters.

In these 6 letters, 'R' occurs 2 times, 'U' occurs 2 times and rest of the letters are different.

Hence, number of ways to arrange these letters

=$\dfrac{6!}{\left(2!\right)\left(2!\right)}$

=$\dfrac{6\times5\times4\times3\times2\times1}{\left(2\times1\right)\left(2\times1\right)}$

=180

44371.How many signals can be made using 6 different coloured flags when any number of them can be hoisted at a time?
1956
1720
2020
1822
Explanation:

Given that any number of flags can be hoisted at a time. Hence we need to find out number of signals that can be made using 1 flag, 2 flags, 3 flags, 4 flags, 5 flags and 6 flags and then add all these.

Number of signals that can be made using 1 flag

= $^6P_{1}$ =6

Number of signals that can be made using 2 flags

=$ ^6P_{2} $

=6×5=30

Number of signals that can be made using 3 flags

=$ ^6P_{3}$

=6×5×4=120

Number of signals that can be made using 4 flags

=$ ^6P_{4} $

=6×5×4×3=360

Number of signals that can be made using 5 flags

= $^6P_{5}$

=6×5×4×3×2=720

Number of signals that can be made using 6 flags

= $^6P_{6 }$

=6×5×4×3×2×1=720

Therefore, required number of signals

=6+30+120+360+720+720=1956

44398.A and B enter in to a partnership and A invests Rs. 10,000 in the partnership. At the end of 4 months he withdraws Rs.2000. At the end of another 5 months, he withdraws another Rs.3000.
If B receives Rs.9600 as his share of the total profit of Rs.19,100 for the year, how much did B invest in the company?
Rs. 12,000
Rs. 8,000
Rs. 6,000
Rs. 96,000
Explanation:
The total profit for the year is 19100. Of this B gets Rs.9600. Therefore, A would get (19100−9600) =Rs. 9500
The partners split their profits in the ratio of their investments.
Therefore, the ratio of the investments of,
A:B=9500:9600=95:96.
A invested Rs.10000 initially for a period of 4 months. Then, he withdrew Rs.2000.
Hence, his investment has reduced to Rs.8000 (for the next 5 months).
Then he withdraws another Rs.3000. Hence, his investment will stand reduced to Rs.5000 during the last three months.
So, the amount of money that he had invested in the company on a money-month basis will be,
=4×10000+5×8000+3×5000
=40000+40000+15000
=95000
If A had 95000 money months invested in the company, B would have had 96,000 money months invested in the company (as the ratio of their investments is 95:96).

If B had 96,000 money-months invested in the company, he has essentially invested $\dfrac{96000}{12}$=Rs. 8000
44405.In a pocket of A, the ratio of Rs.1 coins, 50p coins and 25p coins can be expressed by three consecutive odd prime numbers that are in ascending order. The total value of coins in the bag is Rs 58. If the number of Rs.1, 50p, 25p coins are reversed, find the new total value of coins in the pocket of A?
Rs 68
Rs 43
Rs 75
Rs 82
Explanation:

Since the ratio of the number of Rs. 1, 50p and 25p coins can be represented by 3 consecutive odd numbers that are prime in ascending order, the only possibility for the ratio is 3:5:7.

Let the number of Re1, 50p and 25p coins be 3k,5k and 7k respectively.

Hence, total value of coins in paise

⇒100×3k+50×5k+25×7k
=725k
=5800
⇒k=8.
If the number of coins of Rs. 1,50p and 25p is reversed, the total value of coins in the Bag (in paise)

=100×7k+50×5k+25×3k=1025k (In above we find the value of k).

⇒8200p=Rs. 82.
44415.If $log_{4}x+log_{2}x$=12, then x is equal to:
1024
256
8
16
Explanation:

$log_{4}x+log_{2}x$=12

=>$\dfrac{log\:x}{log\:4}+\dfrac{log\:x}{log\:2}$=12

=>$\dfrac{log\:x}{log\:2^2}+\dfrac{log\:x}{log\:2}$=12

=>$\dfrac{log\:x}{2\:log\:2}+\dfrac{log\:x}{log\:2}$=12

=>$\dfrac{log\:x+2\:log\:x}{2\:log\:2}$=12

=>$\dfrac{3\:log\:x}{2\:log\:2}$=12

Therefore,

logx=$\dfrac{12\times2\:log\:2}{3}$

=8 log 2

=$log(2^8)$

=log(256)

x=256

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