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What should come in place of the question mark (?) in the following question ?
$$\frac{9}{10}+\frac{3}{11}+\frac{7}{15}= ?$$
$$1\frac{217}{330}$$
$$1\frac{221}{330}$$
$$1\frac{211}{330}$$
$$1\frac{197}{330}$$
None of theses
(9/10)+(3/11)+(7/15) = (129/110) + 7/15 = (1/5)(129/22 +7/3) = 1/5(541/66) = 541/330
Converting 541/330 into mixed fraction we get $$1\frac{211}{330}$$
$$\sqrt[3]{42875}-?=21$$
18
13
15
11
None of these
Let unknown be x,
$$\sqrt[3]{42875}+21=x$$
x = 35 - 21 = 14
$$\frac{57}{67}\times\frac{32}{171}\times\frac{45}{128}=?$$
$$\frac{15}{262}$$
$$\frac{15}{268}$$
$$\frac{15}{266}$$
$$\frac{17}{268}$$
The expression, on reframing gives, $$\frac{57}{171}\times\frac{32}{128}\times\frac{45}{67}$$
which on simplifying gives,
(1/4)*(1/3)*(45/67) = (15/268)
$$\frac{1}{4}$$th of $$\frac{1}{2}$$ of $$\frac{3}{4}$$th of 52000=?
4875
4857
4785
4877
In mathematical way, the question is intrepreted as,
(1/4)*(1/2)*(3/4)*52000=(3/32)*52000 = 3*1625 = 4875
$$(7921\div178)-5.5=\sqrt{?}$$
1512
1521
1251
1531
$$(7921\div178)-5.5=\sqrt{x}$$
Hence, x=[(7921/178)-5.5]^2=39^2=1521
38% of 4500-25% of ?=1640
260
270
280
290
Reframing the question by assuming the unknown to be x.
38% of 4500 - 1640 = 25% of x
(x/4) = 1710 - 1640
= 70
x = 280
$$(5863-\sqrt{2704})\times0\cdot5=?$$
2955.5
2905.5
2590.5
2909.5
$$\sqrt{2704} = 52$$
Now,$$( 5863-52)*.5 = 2905.5$$.Therefore, option B is the right answer.
$$(?)^{2}+15^{2}-33^{2}=97$$
33
32
34
30
Rearranging we get, 97+33^2-15^2 = 97 + 1089 - 225 = 961
Now x^2= 961
Hence, x=31
25639-5252-3232=?
17255
17551
17515
17155
The question is written as 25639 - (5252+3232) = 25639 - 8484 = 17155
$$283 \times56+252=20 \times?$$
805
803
807
809
Now, let unknown be x.
283*56=15848
15848+252=16100
20x=16100
x=805
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