Solution 1.
Let the question solved correctly by Mohan = X ;
and number of question solved wrongly by Mohan = 30 - X ;
According to the question,
3X - (30 - X) x 2 = 40
? 3x - 60 + 2x = 40
? 5x - 60 = 40
? 5x = 40 + 60 ? x = 100/5 = 20
? Mohan attempted 20 questions correctly .
Solution 2.
Let the question solved correctly by Mohan = X ;
and number of question solved wrongly by Mohan = Y ;
According to given question;
X + Y = 30 ; -----------i
3X - 2Y = 40 ;-------ii
Now multiply the first equation by 2;
we will get
2X + 2Y = 60 ;--------iii
After Add the equation ii and iii , we will get
3X - 2Y + 2X + 2Y = 40 + 60 ;
? 5X = 100 ;
? X = 20;
? Mohan attempted 20 questions correctly .
Then, | (6x + 6) + 4 | = | 11 |
(5x + 6) + 4 | 10 |
⟹ 10(6x + 10) = 11(5x + 10)
⟹ 5x = 10
⟹ x = 2.
∴ Sagar's present age = (5x + 6) = 16 years.
Let (17.28 ÷ N ) / (3.6 x 0.2) = 2
Then, 17.28 / N = 1.44
? N = 17.28 / 1.44
= 12
? a + (1/b) = 1
ab + 1 = b ......(i)
Also, b + (1/c) = 1
? b = 1 - (1/c) ....(ii)
From Eqs. (i) and (ii), we get
ab + 1 = 1 - (1/c)
? ab = -1/c
? abc = -1
Draw a figure as per given question,
In a Right triangle ADC, Use the (Pythagoras Theorem)
AC = ?DC2 - AD2
? AC = ? 152 - 9 2
? AC = ? 225 - 81
? AC = ? 144
? AC = 12 cm
In a Right triangle BCE, Use the formula
CB = ? CD2 - BE2
? CB = ?15 2 - 122
? CB = ?225 - 144
? CB = ? 81
? CB = 9 m
? Width of the street (AC + BC) = AB = 12 + 9 = 21 m.
Series pattern 4 x 0.5 = 2
2 x 1.5 = 3
Should come in place of 3.5
3 x 2.5 = 7.5
7.5 x 3.5 = 26.25
26.25 x 4.5 = 118.125
Clearly, 3.5 is wrong and is replaced by 3.
Let the number be p and q.
According to question,
(p + q)2 = 4pq
? p2 + q2 + 2pq ? 4pq = 0
? (p ? q)2 = 0
? p = q
Hence , required ratio is 1 : 1 .
According to the question,
4a = 68 [ where a = side]
?a = 68/4 = 17 cm
? Required area = a2
= (17)2 = 289 sq cm
9 | 1 | days |
3 |
12 | 1 | days |
4 |
16 | 1 | days |
3 |
(A's 1 day's work) : (B's 1 day's work) = | 7 | : 1 = 7 : 4. |
4 |
Let A's and B's 1 day's work be 7x and 4x respectively.
Then, 7x + 4x = | 1 | ⟹ 11x = | 1 | ⟹ x = | 1 | . |
7 | 7 | 77 |
∴ A's 1 day's work = | ❨ | 1 | x 7 | ❩ | = | 1 | . |
77 | 11 |
Area = 1/2 hectare = 10000 / 2 m2
= 5000 m2
Again Area = 1/2 x (Diagonal)2
So 1/2 x (Diagonal)2 = 5000m2
? Diagonal2= 10000
? Diagonal = 100
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