Quantitative Aptitude - Arithmetic Ability Questions

Let the large number be x.

Then x - 20 = 20x/100

=> x - x/5 = 20  => x = 25.

Let 4 girls be one unit and now there are 6 units in all.

They can be arranged in 6! ways.

In each of these arrangements 4 girls can be arranged in 4! ways. 

Total number of arrangements in which girls are always together = 6! x 4!= 720 x 24 = 17280

6)4456(742
42
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25
24
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16
12
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4
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Required number = (6-4) = 2.

Let the two parts be Rs. x and Rs. (1301 - x).

 x1+41007=1301-x1+41009

 x1301-x=1+41002=2625*2526

=> 625x=676(1301-x) 

1301x=676 x 1301x=676.
So,the parts are rs.676 and rs.(1301-676)i.e rs.676 and rs.625

The ratio of the ages of A and B is 3 : 5.
The ratio of the ages of B and C is 3 : 5.

B's age is the common link to both these ratio. Therefore, if we make the numerical value of the ratio of B's age in both the ratios same, then we can compare the ages of all 3 in a single ratio.

The can be done by getting the value of B in both ratios to be the LCM of 3 and 5 i.e., 15.

The first ratio between A and B will therefore be 9 : 15 and
the second ratio between B and C will be 15 : 25.

Now combining the two ratios, we get A : B : C = 9 : 15 : 25.

Let their ages be 9x, 15x and 25x.
Then, the sum of their ages will be 9x + 15x + 25x = 49x

The question states that the sum of their ages is 147.
i.e., 49x = 147 or x = 3.

Therefore, B's age = 15x = 15*3 = 45

Height of the triangle = 132-122=25 = 5 cm.

Its area =12*base*height= 12*12*5= 30cm2.