ZS Associates Placement Papers | Online Test Pattern

Let the speed of M be x and the speed of N be y.

∴ Distance covered by M in 2 hrs = 2x and

distance covered by N in 1.5 hrs = 1.5y.

Also, 2x = 1.5(1.5y)

⇒ x : y = 9 : 8.

Ratio in which tea and sugar should be mixed
⇒ 20 – 13 : 13 – 9.5

⇒ 7 : 3.5

⇒ 7 : 3.5

⇒ 2:1 .

Let x be quantity at 9.5/kg.

∴ 2 : 1 = x : 17, hence x = 34 kg.

Greatest six-digit number is 999999.

Divide this number by 12 and get remainder as 3.

Since the remainder is 3, if you subtract 3 from the number, the remaining number will be a multiple of 12.

So the greatest such number will be 999999 – 3 =999996.

Let the total income be x.

Then, income left,

(100 - 80)%of[100-(35 + 25)]%ofx<br /> =20%of40%ofx<br /> =[\frac{20}{100}\times\frac{40}{100)\times100]%ofx<br /> = 8%ofx.

In 18 days, both take the rest of 3 days each and the work, done is \frac{(36-6)}{29} part of the work.

By 18th day \frac{1}{29} part of the work is done more, so it implies that the work was completed on 17th day.

Total time in hours from Sunday 4 pm to the following Sunday at 8 am = [( 6 x 24 ) +16 ] = 160 hours

Thus, the watch gains 6+10\frac{2}{3}=16\frac{2}{3} minutes in 160 hours

therefore, 1 minute is gained in 160\times\frac{3}{50}

⇒6 minutes are gained in

160\times\frac{3}{50}\times6

Required answer = 57\frac{3}{5}hours

Hence, the watch was correct on Wednesday at 1:36 am

Let C.P. be Rs.100

then, S.P. = Rs.133

Let marked price be Rs.x

then, 95% of x = 133

(95/100) * x = 133

x = (133 * 100)/95 = 140

40% above C.P.

Required number of ways

= ⁷C₅*³C₂

=⁷C₂*³C₁

= ((7 * 6)/(2 * 1)) * 3

= 63

We may have 1 black and 2 non-black or 2 black and 1 non-black or 3 black

required number of ways = (³C₁ * ⁶C₂) + (³C₂ * ⁶C₁) + ³C₃

= (3 * ((6 * 5)/(2 * 1))) + (((3 * 2)/(2 * 1)) * 6) + 1

= 45 + 18 + 1 = 64

Consider the ages of 5 children be X, (X + 3), (X + 6), (X + 9) and (X + 12) years

sum of 5 children is 50

then, X + (X + 3) + (X + 6) + (X + 9) + (X + 12) = 50

5X + 30 = 50

5X = 20

X = 4

i.e Age of the youngest child = 4 years.

The logic is 8 × 3 – 5 = 19, 19 × 3 – 5 = 52, 52 × 3 – 5 = 151. So in place of 447, it should be 448.

Daughter of uncle's father — Uncle's sister — Mother;

Mother's son — Brother

Logic is: 190/2 = 95 – 1 = 94
94/2 = 47 -1 = 46
46/2 = 23 – 1 = 22
22/ 2= 11-1= 10
10/2= 5-1= 4
4/2= 2-1 = 1

The arrangement of the persons is
Q W M V N U S O T P R
U is sitting in the center of the row.

The arrangement of the persons is
Q W M V N U S O T P R
OTPR are sitting to the right of S.

The arrangement of the persons is
Q W M V N U S O T P R
Hence N is sitting between V and U.

The arrangement of the persons is
Q W M V N U S O T P R
O and P are the immediate neighbors of T.

The arrangement of the persons is
Q W M V N U S O T P R
If the persons as given in the question change their positions then we have the arrangement P R T V O U S N M Q W
Hence P and W are sitting at the ends.

If the first two statements are true, then Erasers are the most expensive of the three stationery items

Each letter moves +1, -1, … alternately except for L, which is -4. We have to find the word for the code IFWJBP. Add 4 to I then -1, +1 alternately to the remaining letters. The word will be MEXICO.

Required Percentage

=[\frac{(1020 + 1240)}{(8800 + 9500)}\times100]%<br /> =[\frac{2260}{18300}\times100]%<br /> = 12.35%

Required percentage =[\frac{(840+1050+920+980+1020)}{(7500+9200+8450+9200+8800)}\times100]%<br /> =[\frac{4810}{43150}\times100]%<br /> =11.15%

The percentage of candidates qualified to candidates appeared from State P during different years are:

For 1997 (\frac{780}{6400}\times100)%=12.19%

For 1998 (\frac{1020}{8800}\times100)%=11.59%

For 1999 (\frac{890}{7800}\times100)%=11.41%%

For 2000 (\frac{1010}{8750}\times100)%=11.54%

For 2001 (\frac{1250}{9750}\times100)%=12.82%

Therefore, Maximum percentage is for the year 2001

Required average

=[\frac{8100+9500+8700+9700+8950}{5}]=\frac{44950}{5}=8990

Required percentage

=[\frac{(850+920+890+980+1350)}{(7400+8450+7800+8700+9800)}\times100]%<br /> =[\frac{4990}{42150}\times100]%<br /> =11.84%

Total sales of branches B1, B3 and B5 for both the years (in thousand numbers)

= (80 + 105) + (95 + 110) + (75 + 95)

= 560.

Required percentage

=[\frac{(70+80)}{(95+110)}\times100]%<br /> =[\frac{150}{205}\times100]%<br /> =73.17%

Average sales of all the six branches (in thousand numbers) for the year 2000

=1/6 * [80 + 75 + 95 + 85 + 75 + 70]

=80

Required ratio
=[\frac{(75+65)}{(85+95)}]
=\frac{140}{180}
=\frac{7}{9}

Average sales(in thousands) of branches B1, B2, and B6 in 2000

=(\frac{1}{3}\times(80+95+70))[latex]<br /> =[latex](\frac{245}{3})

Average sales(in thousands) of branches B1, B2, and B6 in 2001

=(\frac{1}{3}\times(105+65+110))
=(\frac{280}{3})

therefore, required percentage

=[\frac{\frac{245}{3}}{\frac{280}{3}}\times100]%
=(\frac{245}{280}\times100)%
=87.5%

In the first segment, two letters face right and the next two face left. The first letter in the second segment repeats the last letter of the previous segment. The same is true for the third segment. But the fourth segment changes again; it is the opposite of the first segment, so the last two letters must face right.

Each arrow in this continuing series moves a few degrees in a clockwise direction. Think of these arrows as the big hand on a clock. The first arrow is at noon. The last arrow before the blank would be 12:40. The correct answer is at 12:45.

In each of the segments, the figures alternate between one-half and one-fourth shaded

Notice that in each segment, the figures are all the same shape, but the one in the middle is larger than the two on either side. Also, notice that one of the figures is shaded and that this shading alternates first right and then left. To continue this pattern in the third segment, you will look for a square. Choice 2 is correct because this choice will put the large square between the two smaller squares, with the shading on the right.

Currently there is no PrepInsta Explanation.

Currently there is no PrepInsta Explanation.