Quantitative Aptitude Test - 7

Here n =26, d = 4

Sum = (26/2)*(2a + 25*4)

= 13 [2(z-100) + 100]

= 13 (2z-100)

So, Mean = Sum /26

= 13 (2z-100)/26

= z – 50

Given that k5 < k6 < k7 and n = k5 + k6 + k7

Thus n < 3k7 or k7 > n/3

Therefore, k7 = n/2 or n/1.

But k7 cannot be n/1,

since k5 + k6 ≠ 0,

Thus k7 = n/2.

Now k5 + k6 = n/2

Again, n/2 < 2k6 or k6 > n/4,

thus k6 = n/3.

And then k5 = n/6.

Also, the number n will have total of 8 divisors including 1 and itself (since k7 = n/2 , k8 = n ).

Since n is divisible by three primes 2, 3 and 11, the only possibility for it to have 8 divisors is for it to be the product of the three primes, that is 66.

Option D is correct.

py + 8y = 9

=> y(p + 8) = 9

=> y = 9 / (p + 8)

Here if p = −8 then y will not have a definite value.

So the answer is −8

Hence Option D is correct.

We can split the figure as shown:

3 fully automatic printing machines required 43 days to print a certain number of pages.Therefore, 1 fully automatic printing machine would print 1/(43 × 3) pages in a day.Similarly, 1 semi automatic printing machine would print 1/(43 × 4) pages in a day.7 fully automatic printing machines and 5 semi automatic printing machines would print 7/(43 × 3) + 5/(43 × 4) = 1/12 of the pages in 1 day.Therefore, 7 fully automatic printing machines and 5 semi automatic printing machines would print all the pages in 12 days. Option D is correct.

The speed of the first train = 100/10 = 10 m/s

Relative speed when the trains travelled in opposite directions = (100 + 100)/8 = 25 m/s

Therefore, the speed of the second train = 25 – 10 = 15 m/s

15 × 18/5 = 54 km/h

Distance travelled by the second train in 3 hours = 54 × 3

= 162 km.

Option B is correct.

Total number of arrangements is ⁿP_r where n = 8 and r = 5.
Required number of ways = ⁸P₅

$\frac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{3\times 2\times 2}$

=6720

SP of trader A = 1000 (1 + x).

Profit of trader A = 1000 (1 + x) – 1000.

MP of trader B = 2000 (1 + 2x).

SP of trader B = 2000 (1 + 2x) (1 – x).

Profit of trader B = 2000(1 + 2x) (1 – x) – 2000.

Both make the same profit => 1000(1 + x) – 1000 = 2000(1 + 2x) (1 – x) – 2000

1000x = 2000 – 4000x² + 4000x – 2000x – 2000

4000x² -1000x = 0

1000x (4x – 1) = 0

=> x = 25%

Otherwise % change in imports from the previous year:

2004% Change in import = (150 – 75)*100/75 = 100%

2005% Change in import = (250 – 150)*100/150 = 66.67%

2006% Change in import = (225 – 250)*100/250 = -10%

2007% Change in import = (350 – 225)*100/225 = 55.5%

Option D is correct.

Total imports = 75 + 150 + 250 + 225 + 350 + 275 =1325

Total exports = 150 + 225 + 375 + 300 + 450 + 175 = 1675

Ratio of imports and exports = 1325 : 1675

= 53 : 67.

Option D is correct.